RD Sharma Class 12 Ex 20.2 Solutions Chapter 20 Definite Integrals

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	20
Exercise	20.2
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 20.2 Solutions Chapter 20 Definite Integrals

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

Evaluate the following definite integrals:

Question 1. $\int_{2}^{4}\frac{x}{x^2+1}dx$

Solution:

We have,
I = $\int_{2}^{4}\frac{x}{x^2+1}dx$
I = $\frac{1}{2}\int_{2}^{4}\frac{2x}{x^2+1}dx$
I = $\left[\frac{1}{2}\log(1+x^2)\right]_{2}^{4}$
I = $\frac{1}{2}\log(1+4^2)-\frac{1}{2}\log(1+2^2)$
I = $\frac{1}{2}\log(1+16)-\frac{1}{2}\log(1+4)$
I = $\frac{1}{2}\log17-\frac{1}{2}\log5$
I = $\frac{1}{2}(\log17-log5)$
I = $\frac{1}{2}\log\frac{17}{5}$
Therefore, the value of $\int_{2}^{4}\frac{x}{x^2+1}dx$ is $\frac{1}{2}\log\frac{17}{5}$ .

Question 2. $\int_{1}^{2}\frac{1}{x(1+logx)^2}dx$

Solution:

We have,
I = $\int_{1}^{2}\frac{1}{x(1+logx)^2}dx$
Let 1 + log x = t, so we have,
=> (1/x) dx = 2t dt
Now, the lower limit is, x = 1
=> t = 1 + log x
=> t = 1 + log 1
=> t = 1 + 0
=> t = 1
Also, the upper limit is, x = 2
=> t = 1 + log x
=> t = 1 + log 2
So, the equation becomes,
I = $\int_{1}^{1+\log2}\frac{1}{t^2}dt$
I = $\left[\frac{-1}{t}\right]_1^{1+\log2}$
I = $\left[\frac{-1}{1+\log2}+1\right]$
I = $\frac{-1+1+\log2}{1+\log2}$
I = $\frac{\log2}{1+\log2}$
I = $\frac{\log2}{\log e+\log2}$
I = $\frac{\log2}{\log2e}$
Therefore, the value of $\int_{1}^{2}\frac{1}{x(1+logx)^2}dx$ is $\frac{\log2}{\log2e}$ .

Question 3. $\int_{1}^{2}\frac{3x}{9x^2-1}dx$

Solution:

We have,
I = $\int_{1}^{2}\frac{3x}{9x^2-1}dx$
Let 9x² – 1 = t, so we have,
=> 18x dx = dt
=> 3x dx = dt/6
Now, the lower limit is, x = 1
=> t = 9x² – 1
=> t = 9 (1)² – 1
=> t = 9 – 1
=> t = 8
Also, the upper limit is, x = 2
=> t = 9x² – 1
=> t = 9 (2)² – 1
=> t = 36 – 1
=> t = 35
So, the equation becomes,
I = $\int_{8}^{35}\frac{1}{6t}dt$
I = $\left[\frac{1}{6}\log t\right]_{8}^{35}$
I = $\frac{1}{6}(\log 35-\log 8)$
I = $\frac{1}{6}\log \frac{35}{8}$
Therefore, the value of $\int_{1}^{2}\frac{3x}{9x^2-1}dx$ is $\frac{1}{6}\log \frac{35}{8}$ .

Question 4. $\int_{0}^{\frac{\pi}{2}}\frac{1}{5cosx+3sinx}dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{5cosx+3sinx}dx$
On putting sin x = $\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}$ and cos x = $\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}$ , we get
I = $\int_{0}^{\frac{\pi}{2}}\frac{sec^2\frac{x}{2}}{5(1-tan^2\frac{x}{2})+6tan\frac{x}{2}}dx$
I = $\int_{0}^{\frac{\pi}{2}}\frac{sec^2\frac{x}{2}}{5-5tan^2\frac{x}{2}+6tan\frac{x}{2}}dx$
Let tan x/2 = t. So, we have
=> $\frac{1}{2}sec^2\frac{x}{2}$ = dt
Now, the lower limit is, x = 0
=> t = tan x/2
=> t = tan 0
=> t = 0
Also, the upper limit is, x = π/2
=> t = tan x/2
=> t = tan π/4
=> t = 1
So, the equation becomes,
I = $\int_{0}^{1}\frac{2}{5-5t^2+6t}dt$
I = $\frac{2}{5}\int_{0}^{1}\frac{1}{1-t^2+\frac{6}{5}t}dt$
I = $\frac{2}{5}\int_{0}^{1}\frac{1}{\frac{34}{25}-(t-\frac{3}{5})^2}dt$
I = $\left[\frac{2}{5}.\frac{1}{2}\sqrt{\frac{25}{34}}\log\left(\frac{\sqrt{\frac{34}{25}}+t-\frac{3}{5}}{\sqrt{\frac{34}{25}}-t+\frac{3}{5}}\right)\right]^1_0$
I = $\left[\frac{2}{5}.\frac{1}{2}\frac{5}{\sqrt{34}}\log\left(\frac{\sqrt{\frac{34}{25}}+t-\frac{3}{5}}{\sqrt{\frac{34}{25}}-t+\frac{3}{5}}\right)\right]^1_0$
I = $\frac{1}{\sqrt{34}}\left[\log\left(\frac{\sqrt{34}+2}{\sqrt{34}-2}\right)-\log\left(\frac{\sqrt{34}-3}{\sqrt{34}+3}\right)\right]$
I = $\frac{1}{\sqrt{34}}\log\left(\frac{(\sqrt{34}+2)(\sqrt{34}+3)}{(\sqrt{34}-2)(\sqrt{34}-3)}\right)$
I = $\frac{1}{\sqrt{34}}\log\left(\frac{40+5\sqrt{34}}{40-5\sqrt{34}}\right)$
I = $\frac{1}{\sqrt{34}}\log\left(\frac{8+\sqrt{34}}{8-\sqrt{34}}\right)$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{1}{5cosx+3sinx}dx$ is $\frac{1}{\sqrt{34}}\log\left(\frac{8+\sqrt{34}}{8-\sqrt{34}}\right)$ .

Question 5. $\int_{0}^{a}\frac{x}{\sqrt{a^2+x^2}}dx$

Solution:

We have,
I = $\int_{0}^{a}\frac{x}{\sqrt{a^2+x^2}}dx$
Let a² + x² = t². So, we have
=> 2x dx = 2t dt
=> x dx = t dt
Now, the lower limit is, x = 0
=> t² = a²+ x²
=> t² = a²+ 0²
=> t² = a²
=> t = a
Also, the upper limit is, x = a
=> t² = a² + x²
=> t²= a² + a²
=> t² = 2a²
=> t = √2 a
So, the equation becomes,
I = $\int_{a}^{\sqrt{2}a}\frac{t}{t}dt$
I = $\int_{a}^{\sqrt{2}a}dt$
I = $\left[t\right]_{a}^{\sqrt{2}a}$
I = √2a – a
I = a (√2 – 1)
Therefore, the value of $\int_{0}^{a}\frac{x}{\sqrt{a^2+x^2}}dx$ is a (√2 – 1).

Question 6. $\int_{0}^{1}\frac{e^x}{1+e^{2x}}dx$

Solution:

We have,
I = $\int_{0}^{1}\frac{e^x}{1+e^{2x}}dx$
Let e^x = t. So, we have
=> e^x dx = dt
Now, the lower limit is, x = 0
=> t = e^x
=> t = e⁰
=> t = 1
Also, the upper limit is, x = a
=> t = e^x
=> t = e¹
=> t = e
So, the equation becomes,
I = $\int_{1}^{e}\frac{1}{1+t^2}dt$
I = $\left[tan^{-1}t\right]_{1}^{e}$
I = $tan^{-1}e-tan^{-1}1$
I = $tan^{-1}e-\frac{\pi}{4}$
Therefore, the value of $\int_{0}^{1}\frac{e^x}{1+e^{2x}}dx$ is $tan^{-1}e-\frac{\pi}{4}$ .

Question 7. $\int_{0}^{1}xe^{x^2}dx$

Solution:

We have,
I = $\int_{0}^{1}xe^{x^2}dx$
Let x² = t. So, we have
=> 2x dx = dt
Now, the lower limit is, x = 0
=> t = x²
=> t = 0²
=> t = 0
Also, the upper limit is, x = 1
=> t = x²
=> t = 1²
=> t = 1
So, the equation becomes,
I = $\int_{0}^{1}\frac{e^t}{2}dt$
I = $\frac{1}{2}\int_{0}^{1}e^tdt$
I = $\frac{1}{2}\left[e^t\right]_{0}^{1}$
I = $\frac{1}{2}\left[e^1-e^0\right]$
I = $\frac{1}{2}(e-1)$
Therefore, the value of $\int_{0}^{1}xe^{x^2}dx$ is $\frac{1}{2}(e-1)$ .

Question 8. $\int_{1}^{3}\frac{cos(logx)}{x}dx$

Solution:

We have,
I = $\int_{1}^{3}\frac{cos(logx)}{x}dx$
Let log x = t. So, we have
=> (1/x) dx = dt
Now, the lower limit is, x = 1
=> t = log x
=> t = log 1
=> t = 0
Also, the upper limit is, x = 3
=> t = log x
=> t = log 3
So, the equation becomes,
I = $\int_{0}^{\log3}costdt$
I = $\left[sint\right]_{0}^{\log3}$
I = sin (log 3) – sin 0
I = sin (log 3) – 0
I = sin (log 3)
Therefore, the value of $\int_{1}^{3}\frac{cos(logx)}{x}dx$ is sin (log 3).

Question 9. $\int_{0}^{1}\frac{2x}{1+x^4}dx$

Solution:

We have,
I = $\int_{0}^{1}\frac{2x}{1+x^4}dx$
Let x² = t. So, we have
=> 2x dx = dt
Now, the lower limit is, x = 0
=> t = x²
=> t = 0²
=> t = 0
Also, the upper limit is, x = 1
=> t = x²
=> t = 1²
=> t = 1
So, the equation becomes,
I = $\int_{0}^{1}\frac{1}{1+t^2}dt$
I = $\left[tan^{-1}t\right]_{0}^{1}$
I = $tan^{-1}1-tan^{-1}0$
I = $\frac{\pi}{4}-0$
I = $\frac{\pi}{4}$
Therefore, the value of $\int_{0}^{1}\frac{2x}{1+x^4}dx$ is $\frac{\pi}{4}$ .

Question 10. $\int_{0}^{a}\sqrt{a^2-x^2}dx$

Solution:

We have,
I = $\int_{0}^{a}\sqrt{a^2-x^2}dx$
Let x = a sin t. So, we have
=> dx = a cos t dt
Now, the lower limit is, x = 0
=> a sin t = x
=> a sin t = 0
=> sin t = 0
=> t = 0
Also, the upper limit is, x = a
=> a sin t = a
=> a sin t = a
=> sin t = 1
=> t = π/2
So, the equation becomes,
I = $\int_{0}^{\frac{\pi}{2}}\sqrt{a^2-a^2sin^2t}(acost)dt$
I = $\int_{0}^{\frac{\pi}{2}}\sqrt{a^2(1-sin^2t)}(acost)dt$
I = $\int_{0}^{\frac{\pi}{2}}\sqrt{a^2cos^2t}(acost)dt$
I = $\int_{0}^{\frac{\pi}{2}}(acost)(acost)dt$
I = $\int_{0}^{\frac{\pi}{2}}a^2cos^2tdt$
I = $a^2\int_{0}^{\frac{\pi}{2}}cos^2tdt$
I = $\frac{a^2}{2}\int_{0}^{\frac{\pi}{2}}(1+cos2t)dt$
I = $\frac{a^2}{2}\left[t+\frac{sin2t}{2}\right]_{0}^{\frac{\pi}{2}}$
I = $\frac{a^2}{2}\left[\frac{\pi}{2}+0-0-0\right]$
I = $\frac{a^2}{2}\left[\frac{\pi}{2}\right]$
I = $\frac{\pi a^2}{4}$
Therefore, the value of $\int_{0}^{a}\sqrt{a^2-x^2}dx$ is $\frac{\pi a^2}{4}$ .

Question 11. $\int_{0}^{\frac{\pi}{2}}\sqrt{sinØ}cos^5ØdØ$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}\sqrt{sinØ}cos^5ØdØ$
I = $\int_{0}^{\frac{\pi}{2}}\sqrt{sinØ}cos^4ØcosØdØ$
Let sin Ø = t. So, we have
=> cos Ø dØ = dt
Now, the lower limit is, Ø = 0
=> t = sin Ø
=> t = sin 0
=> t = 0
Also, the upper limit is, Ø = π/2
=> t = sin Ø
=> t = sin π/2
=> t = 1
So, the equation becomes,
I = $\int_{0}^{1}\sqrt{t}(1-t^2)^2dt$
I = $\int_{0}^{1}\sqrt{t}(1+t^4-2t^2)dt$
I = $\int_{0}^{1}(t^\frac{1}{2}+t^\frac{9}{2}-2t^\frac{5}{2})dt$
I = $\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{t^{\frac{9}{2}+1}}{\frac{9}{2}+1}-\frac{2t^{\frac{5}{2}+1}}{\frac{5}{2}+1}\right]_{0}^{1}$
I = $\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{11}{2}}}{\frac{11}{2}}-\frac{2t^{\frac{7}{2}}}{\frac{7}{2}}\right]_{0}^{1}$
I = $\left[\frac{2t^{\frac{3}{2}}}{3}+\frac{2t^{\frac{11}{2}}}{11}-\frac{4t^{\frac{7}{2}}}{7}\right]_{0}^{1}$
I = $\frac{2}{3}+\frac{2}{11}-\frac{4}{7}$
I = $\frac{154+42-132}{231}$
I = $\frac{64}{231}$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\sqrt{sinØ}cos^5ØdØ$ is $\frac{64}{231}$ .

Question 12. $\int_{0}^{\frac{\pi}{2}}\frac{cosx}{1+sinx}dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}\frac{cosx}{1+sin^2x}dx$
Let sin x = t. So, we have
=> cos x dx = dt
Now, the lower limit is, x = 0
=> t = sin x
=> t = sin 0
=> t = 0
Also, the upper limit is, x = π/2
=> t = sin x
=> t = sin π/2
=> t = 1
So, the equation becomes,
I = $\int_{0}^{1}\frac{1}{1+t^2}dt$
I = $\left[tan^{-1}t\right]_{0}^{1}$
I = $tan^{-1}1-tan^{-1}0$
I = $\frac{\pi}{4}-0$
I = $\frac{\pi}{4}$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{cosx}{1+sin^2x}dx$ is $\frac{\pi}{4}$ .

Question 13. $\int_{0}^{\frac{\pi}{2}}\frac{sin\theta}{\sqrt{1+cos\theta}}d\theta$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}\frac{sin\theta}{\sqrt{1+cos\theta}}d\theta$
Let 1 + cos θ = t². So, we have
=> – sin θ dθ = 2t dt
=> sin θ dθ = –2t dt
Now, the lower limit is, θ = 0
=> t² = 1 + cos θ
=> t² = 1 + cos 0
=> t² = 1 + 1
=> t²= 2
=> t = √2
Also, the upper limit is, θ = π/2
=> t² = 1 + cos θ
=> t²= 1 + cos π/2
=> t² = 1 + 0
=> t² = 1
=> t = 1
So, the equation becomes,
I = $\int_{\sqrt{2}}^{1}\frac{-2t}{t}dt$
I = $\int_{\sqrt{2}}^{1}-2dt$
I = $-2\left[t\right]_{\sqrt{2}}^{1}$
I = $-2\left[1-\sqrt{2}\right]$
I = $2\left[\sqrt{2}-1\right]$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{sin\theta}{\sqrt{1+cos\theta}}d\theta$ is $2\left[\sqrt{2}-1\right]$ .

Question 14. $\int_{0}^{\frac{\pi}{3}}\frac{cosx}{3+4sinx}dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{3}}\frac{cosx}{3+4sinx}dx$
Let 3 + 4 sin x = t. So, we have
=> 0 + 4 cos x dx = dt
=> 4 cos x dx = dt
=> cos x dx = dt/4
Now, the lower limit is, x = 0
=> t = 3 + 4 sin x
=> t = 3 + 4 sin 0
=> t = 3 + 0
=> t = 3
Also, the upper limit is, x = π/3
=> t = 3 + 4 sin x
=> t = 3 + 4 sin π/3
=> t = 3 + 4 (√3/2)
=> t = 3 + 2√3
So, the equation becomes,
I = $\int_{3}^{3+2\sqrt{3}}\frac{1}{4t}dt$
I = $\frac{1}{4}\left[\log t\right]_{3}^{3+2\sqrt{3}}$
I = $\frac{1}{4}\left[\log (3+2\sqrt{3})-\log 3)\right]$
I = $\frac{1}{4}\log\frac{3+2\sqrt{3}}{3}$
Therefore, the value of $\int_{0}^{\frac{\pi}{3}}\frac{cosx}{3+4sinx}dx$ is $\frac{1}{4}\log\frac{3+2\sqrt{3}}{3}$ .

Question 15. $\int_{0}^{1}\frac{\sqrt{tan^{-1}x}}{1+x^2}dx$

Solution:

We have,
I = $\int_{0}^{1}\frac{\sqrt{tan^{-1}x}}{1+x^2}dx$
Let tan^–1 x = t. So, we have
=> (1/1+x²) dx = dt
Now, the lower limit is, x = 0
=> t = tan^–1 x
=> t = tan^–1 0
=> t = 0
Also, the upper limit is, x = 1
=> t = tan^–1 t
=> t = tan^–11
=> t = π/4
So, the equation becomes,
I = $\int_{0}^{\frac{\pi}{4}}t^{\frac{1}{2}}dt$
I = $\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{\frac{\pi}{4}}$
I = $\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{\frac{\pi}{4}}$
I = $\frac{2}{3}\left[t^{\frac{3}{2}}\right]_{0}^{\frac{\pi}{4}}$
I = $\frac{2}{3}\left[(\frac{\pi}{4})^{\frac{3}{2}}-0\right]$
I = $\frac{2}{3}(\frac{\pi}{4})^{\frac{3}{2}}$
I = $\frac{2}{24}\pi^{\frac{3}{2}}$
I = $\frac{1}{12}\pi^{\frac{3}{2}}$
Therefore, the value of $\int_{0}^{1}\frac{\sqrt{tan^{-1}x}}{1+x^2}dx$ is $\frac{1}{12}\pi^{\frac{3}{2}}$ .

Question 16. $\int_{0}^{2}x\sqrt{x+2}dx$

Solution:

We have,
I = $\int_{0}^{2}x\sqrt{x+2}dx$
Let x + 2 = t². So, we have
=> dx = 2t dt
Now, the lower limit is, x = 0
=> t² = x + 2
=> t² = 0 + 2
=> t² = 2
=> t = √2
Also, the upper limit is, x = 2
=> t² = x + 2
=> t² = 2 + 2
=> t²= 4
=> t = 2
So, the equation becomes,
I = $\int_{\sqrt{2}}^{2}(t^2-2)\sqrt{t^2}2tdt$
I = $2\int_{\sqrt{2}}^{2}(t^2-2)t^2dt$
I = $2\int_{\sqrt{2}}^{2}(t^4-2t^2)dt$
I = $2\left[\frac{t^5}{5}-\frac{2t^3}{3}\right]_{\sqrt{2}}^{2}$
I = $2\left[\frac{32}{5}-\frac{16}{3}-\frac{4\sqrt{2}}{5}+\frac{4\sqrt{2}}{3}\right]$
I = $2\left[\frac{16+8\sqrt{2}}{15}\right]$
I = $2\left[\frac{8\sqrt{2}(1+\sqrt{2})}{15}\right]$
I = $\frac{16\sqrt{2}(1+\sqrt{2})}{15}$
Therefore, the value of $\int_{0}^{2}x\sqrt{x+2}dx$ is $\frac{16\sqrt{2}(1+\sqrt{2})}{15}$ .

Question 17. $\int_{0}^{1}tan^{-1}\frac{2x}{1-x^2}dx$

Solution:

We have,
I = $\int_{0}^{1}tan^{-1}\frac{2x}{1-x^2}dx$
Let x = tan t. So, we have
=> dx = sec² t dt
Now, the lower limit is, x = 0
=> tan t = x
=> tan x = 0
=> x = 0
Also, the upper limit is, x = 1
=> tan t = x
=> tan x = 1
=> x = π/4
So, the equation becomes,
I = $\int_{0}^{\frac{\pi}{4}}tan^{-1}(\frac{2tant}{1-tan^2t})sec^2tdt$
I = $\int_{0}^{\frac{\pi}{4}}tan^{-1}(tan2t)sec^2tdt$
I = $\int_{0}^{\frac{\pi}{4}}2tsec^2tdt$
I = $2\int_{0}^{\frac{\pi}{4}}tsec^2tdt$
On applying integration by parts method, we get
I = $2\left[t\int_0^\frac{\pi}{4}sec^2tdt-\int_0^\frac{\pi}{4}tantdt\right]$
I = $2\left[ttant-log(cost)\right]^{\frac{\pi}{4}}_0$
I = $2\left[\frac{\pi}{4}+log\frac{1}{\sqrt{2}}-0-0\right]$
I = $2\left[\frac{\pi}{4}+\frac{1}{2}log2\right]$
I = $\frac{\pi}{2}+log2$
Therefore, the value of $\int_{0}^{1}tan^{-1}\frac{2x}{1-x^2}dx$ is $\frac{\pi}{2}+log 2$ .

Question 18. $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{1+sin^4x}dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{1+sin^4x}dx$
Let sin² x = t. So, we have
=> 2 sin x cos x = dt
=> sin x cos x = dt/2
Now, the lower limit is, x = 0
=> t = sin² x
=> t = sin² 0
=> t = 0
Also, the upper limit is, x = π/2
=> t = sin² x
=> t = sin² π/2
=> t = 1
So, the equation becomes,
I = $\int_{0}^{1}\frac{1}{2(1+t^2)}dt$
I = $\frac{1}{2}\int_{0}^{1}\frac{1}{1+t^2}dt$
I = $\frac{1}{2}\left[tan^{-1}t\right]^1_0$
I = $\frac{1}{2}\left[tan^{-1}1-tan^{-1}0\right]$
I = $\frac{1}{2}\left[\frac{\pi}{4}-0\right]$
I = $\frac{1}{2}(\frac{\pi}{4})$
I = $\frac{\pi}{8}$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{1+sin^4x}dx$ is $\frac{\pi}{8}$ .

Question 19. $\int_{0}^{\frac{\pi}{2}}\frac{1}{acosx+bsinx}dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{acosx+bsinx}dx$
On putting cos x = $\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{1-tan^2\frac{x}{2}}{sec^2\frac{x}{2}}$ and sin x = $\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{2tan\frac{x}{2}}{sec^2\frac{x}{2}}$ , we get,
I = $\int_{0}^{\frac{\pi}{2}}\frac{sec^2\frac{x}{2}}{a(1-tan^2\frac{x}{2})+2btan\frac{x}{2}}dx$
Let tan x/2 = t. So, we have
=> 1/2 sec² x/2 dx = dt
=> sec² x/2 dx = 2 dt
Now, the lower limit is, x = 0
=> t = tan x/2
=> t = tan 0/2
=> t = tan 0
=> t = 0
Also, the upper limit is, x = π/2
=> t = tan x/2
=> t = tan π/4
=> t = 1
So, the equation becomes,
I = $2\int_{0}^{1}\frac{1}{a(1-t^2)+2bt}dt$
I = $2\int_{0}^{1}\frac{1}{a-at^2+2bt}dt$
I = $2\int_{0}^{1}\frac{1}{-a(t^2-\frac{2b}{a}t-1)}dt$
I = $\frac{2}{a}\int_{0}^{1}\frac{1}{-\left[(t-\frac{b}{a})^2-1-\frac{b^2}{a^2}\right]}dt$
I = $\frac{2}{a}\int_{0}^{1}\frac{1}{(\frac{b^2}{a^2}+1)-(t-\frac{b}{a})^2}dt$
I = $\left[\frac{2}{a}\left(\frac{1}{2\sqrt{\frac{b^2+a^2}{a^2}}}\right)\log\left(\frac{\sqrt{\frac{b^2+a^2}{a^2}}+(t-\frac{b}{a})}{\sqrt{\frac{b^2+a^2}{a^2}}-(t-\frac{b}{a})}\right)\right]_{0}^{1}$
I = $\frac{1}{\sqrt{b^2+a^2}}\log\left(\frac{a+b+\sqrt{a^2+b^2}}{a+b-\sqrt{a^2+b^2}}\right)$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{1}{acosx+bsinx}dx$ is $\frac{1}{\sqrt{b^2+a^2}}\log\left(\frac{a+b+\sqrt{a^2+b^2}}{a+b-\sqrt{a^2+b^2}}\right)$ .

Question 20. $\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4sinx}dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4sinx}dx$
On putting sin x = $\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}$ , we get
I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4\left(\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}\right)}dx$
I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{5+\left(\frac{8tan\frac{x}{2}}{1+tan^2\frac{x}{2}}\right)}dx$
I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{\frac{5(1+tan^2\frac{x}{2})+8tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}dx$
I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{\frac{5+5tan^2\frac{x}{2}+8tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}dx$
I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{\frac{5+5tan^2\frac{x}{2}+8tan\frac{x}{2}}{sec^2\frac{x}{2}}}dx$
I = $\int_{0}^{\frac{\pi}{2}}\frac{sec^2\frac{x}{2}}{5+5tan^2\frac{x}{2}+8tan\frac{x}{2}}dx$
Let tan x/2 = t. So, we have
=> 1/2 sec² x/2 dx = dt
=> sec²x/2 dx = 2 dt
Now, the lower limit is, x = 0
=> t = tan x/2
=> t = tan 0/2
=> t = tan 0
=> t = 0
Also, the upper limit is, x = π/2
=> t = tan x/2
=> t = tan π/4
=> t = 1
So, the equation becomes,
I = $\int_{0}^{1}\frac{1}{5+5t^2+8t}dt$
I = $\frac{2}{5}\int_{0}^{1}\frac{1}{1+t^2+\frac{8}{5}t}dt$
I = $\frac{2}{5}\int_{0}^{1}\frac{1}{1-\frac{16}{25}+\frac{16}{25}+t^2+\frac{8}{5}t}dt$
I = $\frac{2}{5}\int_{0}^{1}\frac{1}{\frac{9}{25}+\frac{16}{25}+t^2+\frac{8}{5}t}dt$
I = $\frac{2}{5}\int_{0}^{1}\frac{1}{(\frac{3}{5})^2+(t+\frac{4}{5})^2}dt$
I = $\frac{2}{5}\left[\frac{5}{3}tan^{-1}\frac{5}{3}(t+\frac{4}{5})\right]_{0}^{1}$
I = $\frac{2}{3}\left[tan^{-1}(\frac{5t}{3}+\frac{4}{3})\right]_{0}^{1}$
I = $\frac{2}{3}\left[tan^{-1}(\frac{5}{3}+\frac{4}{3})-tan^{-1}(0+\frac{4}{3})\right]$
I = $\frac{2}{3}\left[tan^{-1}(\frac{9}{3})-tan^{-1}(\frac{4}{3})\right]$
I = $\frac{2}{3}\left[tan^{-1}(3)-tan^{-1}(\frac{4}{3})\right]$
I = $\frac{2}{3}\left[tan^{-1}(\frac{3-\frac{4}{3}}{1+3(\frac{4}{3})})\right]$
I = $\frac{2}{3}\left[tan^{-1}(\frac{\frac{5}{3}}{1+4})\right]$
I = $\frac{2}{3}tan^{-1}(\frac{\frac{5}{3}}{5})$
I = $\frac{2}{3}tan^{-1}(\frac{1}{3})$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4sinx}dx$ is $\frac{2}{3}tan^{-1}(\frac{1}{3})$ .

Evaluate the following definite integrals:

Question 21. $\int_{0}^{\pi}\frac{sinx}{sinx+cosx}dx$

Solution:

We have,
I = $\int_{0}^{\pi}\frac{sinx}{sinx+cosx}dx$
Let sin x = A (sin x + cos x) + B $\frac{d}{dx}(sinx+cosx)$
=> sin x = A (sin x + cos x) + B (cos x – sin x)
=> sin x = sin x (A – B) + cos x (A + B)
On comparing both sides, we get
A – B = 1 and A + B = 0
On solving, we get A = 1/2 and B = –1/2.
Therefore, the expression becomes,
I = $\frac{1}{2}\int_0^\pi dx-\frac{1}{2}\int_{0}^{\pi}\frac{cosx-sinx}{sinx+cosx}dx$
I = $\left[\frac{x}{2}\right]_0^\pi-\frac{1}{2}\left[\log(sinx+cosx)\right]_{0}^{\pi}$
I = $\frac{\pi}{2}-\frac{1}{2}(0)$
I = $\frac{\pi}{2}$
Therefore, the value of $\int_{0}^{\pi}\frac{sinx}{sinx+cosx}dx$ is $\frac{\pi}{2}$ .

Question 22. $\int_{0}^{\pi}\frac{1}{3+2sinx+cosx}dx$

Solution:

We have,
I = $\int_{0}^{\pi}\frac{1}{3+2sinx+cosx}dx$
On putting cos x = $\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}$ and sin x = $\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}$ , we get,
I = $\int_{0}^{\pi}\frac{1}{3+\frac{4tan\frac{x}{2}}{1+tan^2\frac{x}{2}}+\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}}dx$
I = $\int_{0}^{\pi}\frac{1+tan^2\frac{x}{2}}{3(1+tan^2\frac{x}{2})+4tan\frac{x}{2}+1-tan^2\frac{x}{2}}dx$
I = $\int_{0}^{\pi}\frac{sec^2\frac{x}{2}}{3+3tan^2\frac{x}{2}+4tan\frac{x}{2}+1-tan^2\frac{x}{2}}dx$
I = $\int_{0}^{\pi}\frac{sec^2\frac{x}{2}}{2tan^2\frac{x}{2}+4tan\frac{x}{2}+4}dx$
Let tan x/2 = t. So, we have
=> 1/2 sec² x/2 dx = dt
=> sec² x/2 dx = 2 dt
Now, the lower limit is, x = 0
=> t = tan x/2
=> t = tan 0/2
=> t = tan 0
=> t = 0
Also, the upper limit is, x = π
=> t = tan x/2
=> t = tan π/2
=> t = ∞
So, the equation becomes,
I = $\int_{0}^{\infty}\frac{2}{2t^2+4t+4}dt$
I = $\int_{0}^{\infty}\frac{1}{t^2+2t+2}dt$
I = $\int_{0}^{\infty}\frac{1}{(t+1)^2+1}dt$
I = $\left[tan^{-1}(t+1)\right]^\infty_0$
I = $tan^{-1}\infty-tan^{-1}1$
I = $\frac{\pi}{2}-\frac{\pi}{4}$
I = $\frac{\pi}{4}$
Therefore, the value of $\int_{0}^{\pi}\frac{1}{3+2sinx+cosx}dx$ is $\frac{\pi}{4}$ .

Question 23. $\int_{0}^{1}tan^{-1}xdx$

Solution:

We have,
I = $\int_{0}^{1}tan^{-1}xdx$
I = $tan^{-1}x\int_{0}^{1}dx-\int_{0}^1(\int dx) \frac{d}{dx}(tan^{-1}x)dx$
I = $\left[xtan^{-1}x\right]_{0}^{1}-\int_{0}^1\frac{x}{1+x^2}dx$
I = $\left[xtan^{-1}x\right]_{0}^{1}-\frac{1}{2}\int_{0}^1\frac{2x}{1+x^2}dx$
I = $\left[xtan^{-1}x\right]_{0}^{1}-\frac{1}{2}\left[log(1+x^2)\right]_{0}^1$
I = $(\frac{\pi}{4}-0)-\frac{1}{2}\left[log(1+1)-log(1+0)\right]$
I = $\frac{\pi}{4}-\frac{1}{2}\left[log2-log1\right]$
I = $\frac{\pi}{4}-\frac{1}{2}\left[log2-0\right]$
I = $\frac{\pi}{4}-\frac{\log2}{2}$
Therefore, the value of $\int_{0}^{1}tan^{-1}xdx$ is $\frac{\pi}{4}-\frac{\log2}{2}$ .

Question 24. $\int_{0}^{\frac{1}{2}}\frac{xsin^{-1}x}{\sqrt{1-x^2}}dx$

Solution:

We have,
I = $\int_{0}^{\frac{1}{2}}\frac{xsin^{-1}x}{\sqrt{1-x^2}}dx$
Let sin^–1 x = t. So, we have
=> $\frac{1}{\sqrt{1+x^2}}dx$ = dt
Now, the lower limit is, x = 0
=> t = sin^–1 x
=> t = sin^–1 0
=> t = 0
Also, the upper limit is, x = 1/2
=> t = sin^–1 x
=> t = sin^–1 1/2
=> t = π/6
So, the equation becomes,
I = $\int_{0}^{\frac{\pi}{6}}tsintdt$
I = $t\int_{0}^{\frac{\pi}{6}}sintdt-\int_{0}^{\frac{\pi}{6}}(\int sintdt) \frac{d}{dt}(t)dt$
I = $\left[-tcost\right]_{0}^{\frac{\pi}{6}}+\int_{0}^{\frac{\pi}{6}}costdt$
I = $\left[-tcost\right]_{0}^{\frac{\pi}{6}}+\left[sint\right]_{0}^{\frac{\pi}{6}}$
I = $\left[-\frac{\pi}{6}(\frac{\sqrt{3}}{2})+0\right]+\left[sin\frac{\pi}{6}-0\right]$
I = $\frac{-\sqrt{3}\pi}{12}+\frac{1}{2}$
I = $\frac{1}{2}-\frac{\sqrt{3}\pi}{12}$
Therefore, the value of $\int_{0}^{\frac{1}{2}}\frac{xsin^{-1}x}{\sqrt{1-x^2}}dx$ is $\frac{1}{2}-\frac{\sqrt{3}\pi}{12}$ .

Question 25. $\int_{0}^{\frac{\pi}{4}}(\sqrt{tanx}+\sqrt{cotx})dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{4}}(\sqrt{tanx}+\sqrt{cotx})dx$
I = $\int_{0}^{\frac{\pi}{4}}(\sqrt{\frac{sinx}{cosx}}+\sqrt{\frac{cosx}{sinx}})dx$
I = $\int_{0}^{\frac{\pi}{4}}(\frac{sinx+cosx}{\sqrt{sinxcosx}})dx$
I = $\sqrt{2}\int_{0}^{\frac{\pi}{4}}(\frac{sinx+cosx}{\sqrt{2sinxcosx}})dx$
I = $\sqrt{2}\int_{0}^{\frac{\pi}{4}}(\frac{sinx+cosx}{\sqrt{1-(sinx-cosx)^2}})dx$
Let sinx – cosx = t. So, we have
=> (cos x + sin x) dx = dt
Now, the lower limit is, x = 0
=> t = sinx – cosx
=> t = sin 0 – cos 0
=> t = 0 – 1
=> t = –1
Also, the upper limit is, x = π/4
=> t = sinx – cosx
=> t = sin π/4 – cos π/4
=> t = sin π/4 – sin π/4
=> t = 0
So, the equation becomes,
I = $\sqrt{2}\int_{-1}^{0}\frac{1}{\sqrt{1-t^2}}dt$
I = $\sqrt{2}\left[sin^{-1}t\right]_{-1}^{0}$
I = $\sqrt{2}\left[sin^{-1}0-sin^{-1}(-1)\right]$
I = $\sqrt{2}\left[0+sin^{-1}(1)\right]$
I = $\sqrt{2}(\frac{\pi}{2})$
I = $\frac{\pi}{\sqrt{2}}$
Therefore, the value of $\int_{0}^{\frac{\pi}{4}}(\sqrt{tanx}+\sqrt{cotx})dx$ is $\frac{\pi}{\sqrt{2}}$ .

Question 26. $\int_{0}^{\frac{\pi}{4}}\frac{tan^3x}{1+cos2x}dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{4}}\frac{tan^3x}{1+cos2x}dx$
I = $\int_{0}^{\frac{\pi}{4}}\frac{tan^3x}{2cos^2x}dx$
I = $\int_{0}^{\frac{\pi}{4}}\frac{tan^3xsec^2x}{2}dx$
I = $\frac{1}{2}\int_{0}^{\frac{\pi}{4}}tan^3xsec^2xdx$
Let tan x = t. So, we have
=> sec² x dx = dt
Now, the lower limit is, x = 0
=> t = tan x
=> t = tan 0
=> t = 0
Also, the upper limit is, x = π/4
=> t = tan x
=> t = tan π/4
=> t = 1
So, the equation becomes,
I = $\frac{1}{2}\int_{0}^{1}t^3dt$
I = $\frac{1}{2}\left[\frac{t^4}{4}\right]_{0}^{1}$
I = $\frac{1}{2}(\frac{1}{4}-0)$
I = $\frac{1}{8}$
Therefore, the value of $\int_{0}^{\frac{\pi}{4}}\frac{tan^3x}{1+cos2x}dx$ is $\frac{1}{8}$ .

Question 27. $\int_{0}^{\pi}\frac{1}{5+3cosx}dx$

Solution:

We have,
I = $\int_{0}^{\pi}\frac{1}{5+3cosx}dx$
On putting cos x = $\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}$ , we get
I = $\int_{0}^{\pi}\frac{1}{5+\frac{3(1-tan^2\frac{x}{2})}{1+tan^2\frac{x}{2}}}dx$
I = $\int_{0}^{\pi}\frac{1+tan^2\frac{x}{2}}{5(1+tan^2\frac{x}{2})+3(1-tan^2\frac{x}{2})}dx$
I = $\int_{0}^{\pi}\frac{sec^2\frac{x}{2}}{5(1+tan^2\frac{x}{2})+3(1-tan^2\frac{x}{2})}dx$
Let tan x/2 = t. So, we have
=> 1/2 sec² x/2 dx = dt
=> sec² x/2 dx = 2 dt
Now, the lower limit is, x = 0
=> t = tan x/2
=> t = tan 0/2
=> t = tan 0
=> t = 0
Also, the upper limit is, x = π
=> t = tan x/2
=> t = tan π/2
=> t = ∞
So, the equation becomes,
I = $\int_{0}^{\infty}\frac{1}{5(1+t^2)+3(1-t^2)}dt$
I = $\int_{0}^{\infty}\frac{1}{5+5t^2+3-3t^2}dt$
I = $\int_{0}^{\infty}\frac{1}{8+2t^2}dt$
I = $\frac{1}{2}\int_{0}^{\infty}\frac{1}{4+t^2}dt$
I = $\frac{1}{2}\left[tan^{-1}\frac{t}{2}\right]_{0}^{\infty}$
I = $\frac{1}{2}\left[tan^{-1}\frac{\infty}{2}-\tan^{-1}\frac{0}{2}\right]$
I = $\frac{1}{2}\left[tan^{-1}\infty-\tan^{-1}0\right]$
I = $\frac{1}{2}(\frac{\pi}{2})$
I = $\frac{\pi}{4}$
Therefore, the value of $\int_{0}^{\pi}\frac{1}{5+3cosx}dx$ is $\frac{\pi}{4}$ .

Question 28. $\int_{0}^{\frac{\pi}{2}}\frac{1}{a^2sin^2x+b^2cos^2x}dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{a^2sin^2x+b^2cos^2x}dx$
I = $\int_{0}^{\frac{\pi}{2}}\frac{\frac{1}{cos^2x}}{a^2\frac{sin^2x}{cos^2x}+b^2\frac{cos^2x}{cos^2x}}dx$
I = $\int_{0}^{\frac{\pi}{2}}\frac{sec^2x}{a^2tan^2x+b^2}dx$
I = $\frac{1}{a^2}\int_{0}^{\frac{\pi}{2}}\frac{sec^2x}{tan^2x+\frac{b^2}{a^2}}dx$
Let tan x = t. So, we have
=> sec²x dx = dt
Now, the lower limit is, x = 0
=> t = tan x
=> t = tan 0
=> t = 0
Also, the upper limit is, x = π/2
=> t = tan x
=> t = tan π/2
=> t = ∞
So, the equation becomes,
I = $\frac{1}{a^2}\int_{0}^{\infty}\frac{1}{t^2+\frac{b^2}{a^2}}dt$
I = $\frac{1}{a^2}\left[\frac{a}{b}tan^{-1}\frac{at}{b}\right]_{0}^{\infty}$
I = $\frac{1}{a^2}\left[\frac{a}{b}tan^{-1}\infty-\frac{a}{b}tan^{-1}0\right]$
I = $\frac{1}{a^2}(\frac{a}{b})(\frac{\pi}{2})$
I = $\frac{\pi}{2ab}$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{1}{a^2sin^2x+b^2cos^2x}dx$ is $\frac{\pi}{2ab}$ .

Question 29. $\int_{0}^{\frac{\pi}{2}}\frac{x+sinx}{1+cosx}dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}\frac{x+sinx}{1+cosx}dx$
I = $\int_{0}^{\frac{\pi}{2}}\frac{x+2sin\frac{x}{2}cos\frac{x}{2}}{2cos^2\frac{x}{2}}dx$
I = $\int_{0}^{\frac{\pi}{2}}\frac{xsec^2\frac{x}{2}}{2}+tan\frac{x}{2}dx$
I = $\left[xtan\frac{x}{2}-\int_{0}^{\frac{\pi}{2}}tan\frac{x}{2}dx+\int_{0}^{\frac{\pi}{2}}tan\frac{x}{2}dx\right]^{\frac{\pi}{2}}_0$
I = $\left[xtan\frac{x}{2}\right]^\frac{\pi}{2}_0$
I = $\frac{\pi}{2}tan\frac{\pi}{4}-0$
I = $\frac{\pi}{2}$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{x+sinx}{1+cosx}dx$ is $\frac{\pi}{2}$ .

Question 30. $\int_{0}^{1}\frac{tan^{-1}x}{1+x^2}dx$

Solution:

We have,
I = $\int_{0}^{1}\frac{tan^{-1}x}{1+x^2}dx$
Let tan^–1 x = t. So, we have
=> $\frac{1}{1+x^2}dx$ = dt
Now, the lower limit is, x = 0
=> t = tan^–1 x
=> t = tan^–1 0
=> t = 0
Also, the upper limit is, x = 1
=> t = tan^–1 x
=> t = tan^–1 1
=> t = π/4
So, the equation becomes,
I = $\int_{0}^{\frac{\pi}{4}}tdt$
I = $\left[\frac{t^2}{2}\right]_{0}^{\frac{\pi}{4}}$
I = $\frac{1}{2}(\frac{\pi^2}{16})-0$
I = $\frac{\pi^2}{32}$
Therefore, the value of $\int_{0}^{1}\frac{tan^{-1}x}{1+x^2}dx$ is $\frac{\pi^2}{32}$ .

Question 31. $\int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{3+sin2x}dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{3+sin2x}dx$
I = $\int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{3+1-(cosx-sinx)^2}dx$
I = $\int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{4-(cosx-sinx)^2}dx$
I = $\left[\frac{1}{4}log|\frac{2+sinx-cosx}{2-sinx+cosx}|\right]_{0}^{\frac{\pi}{4}}$
I = $\frac{1}{4}\left[log|\frac{2+sin\frac{\pi}{4}-cos\frac{\pi}{4}}{2-sin\frac{\pi}{4}+cos\frac{\pi}{4}}|-log\frac{2+0-1}{2-0+1}\right]$
I = $\frac{1}{4}\left(log|\frac{2}{2}|-log\frac{1}{3}\right)$
I = $\frac{1}{4}\left(log1-log\frac{1}{3}\right)$
I = $\frac{1}{4}\left(-log\frac{1}{3}\right)$
I = $-\frac{log\frac{1}{3}}{4}$
I = $\frac{log3}{4}$
Therefore, the value of $\int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{3+sin2x}dx$ is $\frac{log3}{4}$ .

Question 32. $\int_{0}^{1}xtan^{-1}xdx$

Solution:

We have,
I = $\int_{0}^{1}xtan^{-1}xdx$
On using integration by parts, we get
I = $tan^{-1}x\int_{0}^{1}xdx-\int_0^1(\int xdx)\frac{d}{dx}(tan^{-1}x)dx$
I = $\left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_0^1\frac{x^2}{1+x^2}dx$
I = $\left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_0^1\frac{1+x^2-1}{1+x^2}dx$
I = $\left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_0^1\frac{1+x^2}{1+x^2}dx+\frac{1}{2}\int_0^1\frac{1}{1+x^2}dx$
I = $\left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_0^1\frac{1+x^2}{1+x^2}dx+\frac{1}{2}\int_0^1\frac{1}{1+x^2}dx$
I = $\left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_0^1dx+\frac{1}{2}\int_0^1\frac{1}{1+x^2}dx$
I = $\left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\left[x\right]_0^1+\frac{1}{2}\left[tan^{-1}x\right]_0^1$
I = $(\frac{\pi}{8}-0)-\frac{1}{2}(1-0)+\frac{1}{2}\left[tan^{-1}1-tan^{-1}0\right]$
I = $\frac{\pi}{8}-\frac{1}{2}+\frac{1}{2}(\frac{\pi}{4}-0)$
I = $\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}$
I = $\frac{\pi}{4}-\frac{1}{2}$
Therefore, the value of $\int_{0}^{1}xtan^{-1}xdx$ is $\frac{\pi}{4}-\frac{1}{2}$ .

Question 33. $\int_{0}^{1}\frac{1-x^2}{x^4+x^2+1}dx$

Solution:

We have,
I = $\int_{0}^{1}\frac{1-x^2}{x^4+x^2+1}dx$
I = $-\int_{0}^{1}\frac{x^2-1}{x^4+x^2+1}dx$
I = $-\int_{0}^{1}\frac{x^2(1-\frac{1}{x^2})}{x^2(x^2+1+\frac{1}{x^2})}dx$
I = $-\int_{0}^{1}\frac{1-\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}dx$
I = $-\int_{0}^{1}\frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})^2-1^2}dx$
I = $-\left[\frac{1}{2}\log|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}|\right]^1_0$
I = $-\left[\frac{1}{2}\log|\frac{x^2-x+1}{x^2+x+1}|\right]^1_0$
I = $-\frac{1}{2}\log|\frac{1^2-1+1}{1^2+1+1}|+\frac{1}{2}\log|\frac{0^2-0+1}{0^2+0+1}|$
I = $-\frac{1}{2}\log\frac{1}{3}+\frac{1}{2}\log1$
I = $-\frac{1}{2}\log\frac{1}{3}+\frac{1}{2}(0)$
I = $-\frac{1}{2}\log\frac{1}{3}$
I = $\log(\frac{1}{3})^{\frac{-1}{2}}$
I = $\log3^{\frac{1}{2}}$
I = $\frac{1}{2}\log3$
Therefore, the value of $\int_{0}^{1}\frac{1-x^2}{x^4+x^2+1}dx$ is $\frac{1}{2}\log3$ .

Question 34. $\int_{0}^{1}\frac{24x^3}{(1+x^2)^4}dx$

Solution:

We have,
I = $\int_{0}^{1}\frac{24x^3}{(1+x^2)^4}dx$
Let 1 + x² = t. So, we have
=> 2x dx = dt
Now, the lower limit is, x = 0
=> t = 1 + x²
=> t = 1 + 0²
=> t = 1 + 0
=> t = 1
Also, the upper limit is, x = π
=> t = 1 + x²
=> t = 1 + 1²
=> t = 1 + 1
=> t = 2
So, the equation becomes,
I = $\int_{0}^{2}\frac{12(2x)(x^2)}{(1+x^2)^4}dx$
I = $\int_{1}^{2}\frac{12(t-1)}{t^4}dt$
I = $12\int_{1}^{2}(\frac{t}{t^4}-\frac{1}{t^4})dt$
I = $12\int_{1}^{2}(\frac{1}{t^3}-\frac{1}{t^4})dt$
I = $12\left[\frac{-1}{2t^2}-\frac{1}{3t^3}\right]_{1}^{2}$
I = $12\left[\frac{-1}{8}+\frac{1}{24}+\frac{1}{2}-\frac{1}{3}\right]$
I = $12\left[\frac{-3+1+12-8}{24}\right]$
I = $12(\frac{2}{24})$
I = 1
Therefore, the value of $\int_{0}^{1}\frac{24x^3}{(1+x^2)^4}dx$ is 1.

Question 35. $\int_{4}^{12}x(x-4)^{\frac{1}{3}}dx$

Solution:

We have,
I = $\int_{4}^{12}x(x-4)^{\frac{1}{3}}dx$
Let x – 4 = t³. So, we have
=> dx = 3t² dt
Now, the lower limit is, x = 4
=> t³ = x – 4
=> t³ = 4 – 4
=> t³ = 0
=> t = 0
Also, the upper limit is, x = 12
=> t³ = x – 4
=> t³ = 12 – 4
=> t³ = 8
=> t = 2
So, the equation becomes,
I = $\int_{0}^{2}(t^3+4)t(3t^2)dt$
I = $\int_{0}^{2}3t^3(t^3+4)dt$
I = $3\int_{0}^{2}(t^6+4t^3)dt$
I = $3\left[\frac{t^7}{7}+t^4\right]_{0}^{2}$
I = $3\left[\frac{128}{7}+16-0-0\right]$
I = $\frac{720}{7}$
Therefore, the value of $\int_{4}^{12}x(x-4)^{\frac{1}{3}}dx$ is $\frac{720}{7}$ .

Question 36. $\int_{0}^{\frac{\pi}{2}}x^2sinxdx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}x^2sinxdx$
On using integration by parts, we get
I = $x^2\int sinxdx-\int(\int sinxdx)\frac{d}{dx}(x^2)dx$
I = $x^2cosx-\int 2xcosxdx$
I = $x^2cosx+2[x\int cosxdx-\int(\int cosxdx)\frac{dx}{dx}dx]$
I = $x^2cosx+2[xsinxdx-\int sinxdx]$
I = $\left[x^2cosx+2xsinxdx+2cosx\right]_0^{\frac{\pi}{2}}$
I = π + 0 – 0 – 0 – 2
I = π – 2
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}x^2sinxdx$ is π – 2.

Question 37. $\int_{0}^{1}\sqrt{\frac{1-x}{1+x}}dx$

Solution:

We have,
I = $\int_{0}^{1}\sqrt{\frac{1-x}{1+x}}dx$
Let x = cos 2t. So, we have
=> dx = – 2 sin 2t dt
Now, the lower limit is, x = 0
=> cos 2t = x
=> cos 2t = 0
=> 2t = π/2
=> t = π/4
Also, the upper limit is, x = 1
=> cos 2t = x
=> cos 2t = 1
=> 2t = 0
=> t = 0
So, the equation becomes,
I = $\int_{\frac{\pi}{4}}^{0}\sqrt{\frac{1-cos2t}{1+cos2t}}(-2sin2t)dt$
I = $2\int^{\frac{\pi}{4}}_{0}\sqrt{\frac{sin^2t}{cos^2t}}(sin2t)dt$
I = $2\int^{\frac{\pi}{4}}_{0}\frac{sint}{cost}(2sintcost)dt$
I = $4\int^{\frac{\pi}{4}}_{0}sin^2tdt$
I = $2\int^{\frac{\pi}{4}}_{0}2sin^2tdt$
I = $2\int^{\frac{\pi}{4}}_{0}(1-cos2t)dt$
I = $2\left[t-\frac{sin^2t}{2}\right]^{\frac{\pi}{4}}_{0}$
I = $2\left[\frac{\pi}{4}-\frac{1}{2}\right]$
I = $\frac{\pi}{2}-1$
Therefore, the value of $\int_{0}^{1}\sqrt{\frac{1-x}{1+x}}dx$ is $\frac{\pi}{2}-1$ .

Question 38. $\int_{0}^{1}\frac{1-x^2}{(1+x^2)^2}dx$

Solution:

We have,
I = $\int_{0}^{1}\frac{1-x^2}{(1+x^2)^2}dx$
I = $\int_{0}^{1}\frac{-x^2(1-\frac{1}{x^2})}{x^2(x+\frac{1}{x})^2}dx$
I = $-\int_{0}^{1}\frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})^2}dx$
Let x + 1/x = t. So, we have
=> (1 – 1/x²)dx = dt
Now, the lower limit is, x = 0
=> t = x + 1/x
=> t = ∞
Also, the upper limit is, x = 1
=> t = x + 1/x
=> t = 1 + 1
=> t = 2
So, the equation becomes,
I = $-\int_{\infty}^{2}\frac{dt}{t^2}$
I = $\int^{\infty}_{2}\frac{dt}{t^2}$
I = $\left[\frac{-1}{t}\right]^{\infty}_{2}$
I = $\frac{-1}{\infty}+\frac{1}{2}$
I = $\frac{1}{2}$
Therefore, the value of $\int_{0}^{1}\frac{1-x^2}{(1+x^2)^2}dx$ is $\frac{1}{2}$ .

Question 39. $\int_{-1}^{1}5x^4\sqrt{x^2+1}dx$

Solution:

We have,
I = $\int_{-1}^{1}5x^4\sqrt{x^2+1}dx$
Let x⁵ + 1 = t. So, we have
=> 5x⁴ dx = dt
Now, the lower limit is, x = –1
=> t = x⁵ + 1
=> t = (–1)⁵ + 1
=> t = –1 + 1
=> t = 0
Also, the upper limit is, x = 1
=> t = x⁵ + 1
=> t = (1)⁵ + 1
=> t = 1 + 1
=> t = 2
So, the equation becomes,
I = $\int_{0}^{2}\sqrt{t}dt$
I = $\left[\frac{2}{3}t^{\frac{3}{2}}\right]_{0}^{2}$
I = $\frac{2}{3}(2^{\frac{3}{2}})$
I = $\frac{2}{3}(2\sqrt{2})$
I = $\frac{4\sqrt{2}}{3}$
Therefore, the value of $\int_{-1}^{1}5x^4\sqrt{x^2+1}dx$ is $\frac{4\sqrt{2}}{3}$ .

Question 40. $\int_{0}^{\frac{\pi}{2}}\frac{cos^2x}{1+3sin^2x}dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}\frac{cos^2x}{1+3sin^2x}dx$
I = $\int_{0}^{\frac{\pi}{2}}\frac{sec^2x}{sec^2x(sec^2x+3tan^2x)}dx$
Let tan x = t. So, we have
=> sec²x dx = dt
Now, the lower limit is, x = 0
=> t = tan x
=> t = tan 0
=> t = 0
Also, the upper limit is, x = π/2
=> t = tan x
=> t = tan π/2
=> t = ∞
So, the equation becomes,
I = $\int_{0}^{\infty}\frac{1}{(1+t^2)(1+4t^2)}dx$
I = $\frac{-1}{3}\int_{0}^{\infty}(\frac{1}{1+t^2}-\frac{1}{1+4t^2})dt$
I = $\frac{-1}{3}\left[tan^{-1}t-2tan^{-1}2t\right]_{0}^{\infty}$
I = $\frac{-1}{3}(\frac{\pi}{2})$
I = $\frac{-\pi}{6}$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{cos^2x}{1+3sin^2x}dx$ is $\frac{-\pi}{6}$ .

Question 41. $\int_{0}^{\frac{\pi}{4}}sin^32tcos2tdt$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{4}}sin^32tcos2tdt$
Let sin 2t = u. So, we have
=> 2 cos 2t dt = du
=> cos 2t dt = du/2
Now, the lower limit is, x = 0
=> u = sin 2t
=> u = sin 0
=> u = 0
Also, the upper limit is, x = π/4
=> u = sin 2t
=> u = sin π/2
=> u = 1
So, the equation becomes,
I = $\frac{1}{2}\int_{0}^{1}u^3du$
I = $\frac{1}{2}\left[\frac{u^4}{4}\right]_{0}^{1}$
I = $\frac{1}{2}(\frac{1}{4})$
I = $\frac{1}{8}$
Therefore, the value of $\int_{0}^{\frac{\pi}{4}}sin^32tcos2tdt$ is $\frac{1}{8}$ .

Evaluate the following definite integrals:

Question 42. $\int_{0}^{\pi}5(5-4\cos\theta)^{\frac{1}{4}}\sin\theta d\theta$

Solution:

We have,
I = $\int_{0}^{\pi}5(5-4\cos\theta)^{\frac{1}{4}}\sin\theta d\theta$
Let 5 – 4 cos θ = t. So, we have
=> 4 sin θ dθ = dt
=> sin θ dθ = dt/4
Now, the lower limit is, θ = 0
=> t = 5 – 4 cos θ
=> t = 5 – 4 cos 0
=> t = 5 – 4
=> t = 1
Also, the upper limit is, θ = π
=> t = 5 – 4 cos θ
=> t = 5 – 4 cos π
=> t = 5 + 4
=> t = 9
So, the equation becomes,
I = $\int_{1}^{9}\frac{5t^{\frac{1}{4}}}{4}dt$
I = $\frac{5}{4}\int_{1}^{9}t^{\frac{1}{4}}dt$
I = $\frac{5}{4}\left[\frac{t^{\frac{5}{4}}}{\frac{5}{4}}\right]_{1}^{9}$
I = $\frac{5}{4}\left[\frac{4}{5}t^{\frac{5}{4}}\right]_{1}^{9}$
I = $\left[t^{\frac{5}{4}}\right]_{1}^{9}$
I = $9^{\frac{5}{4}}-1^{\frac{5}{4}}$
I = $3^{\frac{5}{2}}-1$
I = 9√3 – 1
Therefore, the value of $\int_{0}^{\pi}5(5-4\cos\theta)^{\frac{1}{4}}\sin\theta d\theta$ is 9√3 – 1.

Question 43. $\int_{0}^{\frac{\pi}{6}}\cos^{-3}2\theta \sin2\theta d\theta$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{6}}\cos^{-3}2\theta \sin2\theta d\theta$
I = $\int_{0}^{\frac{\pi}{6}}\frac{\sin2\theta}{\cos^{3}2\theta} d\theta$
I = $\int_{0}^{\frac{\pi}{6}}\frac{\sin2\theta}{(\cos2\theta)(\cos^22\theta)} d\theta$
I = $\int_{0}^{\frac{\pi}{6}}\frac{\tan2\theta}{\cos^22\theta} d\theta$
I = $\int_{0}^{\frac{\pi}{6}}\tan2\theta \sec^22\theta d\theta$
Let tan 2θ = t. So, we have
=> 2 sec² 2θ dθ = dt
=> sec² 2θ dθ = dt/2
Now, the lower limit is, θ = 0
=> t = tan 2θ
=> t = tan 0
=> t = 0
Also, the upper limit is, θ = π/6
=> t = tan 2θ
=> t = tan π/3
=> t = √3
So, the equation becomes,
I = $\frac{1}{2}\int_{0}^{\sqrt{3}}tdt$
I = $\frac{1}{2}\left[\frac{t^2}{2}\right]_{0}^{\sqrt{3}}$
I = $\frac{1}{2}\left[\frac{3}{2}-0\right]$
I = $\frac{3}{4}$
Therefore, the value of $\int_{0}^{\frac{\pi}{6}}\cos^{-3}2\theta \sin2\theta d\theta$ is $\frac{3}{4}$ .

Question 44. $\int_{0}^{\pi^\frac{3}{2}}\sqrt{x}cos^2x^\frac{2}{3}dx$

Solution:

We have,
I = $\int_{0}^{\pi^\frac{3}{2}}\sqrt{x}cos^2x^\frac{2}{3}dx$
Let $x^{\frac{2}{3}}$ = t. So, we have
=> $\frac{3}{2}\sqrt{x}dx$ = dt
Now, the lower limit is, x = 0
=> t = $x^{\frac{2}{3}}$
=> t = $0^{\frac{2}{3}}$
=> t = 0
Also, the upper limit is, x = $\pi^{\frac{3}{2}}$
=> t = $x^{\frac{2}{3}}$
=> t = $(\pi^{\frac{3}{2}})^{\frac{2}{3}}$
=> t = π
So, the equation becomes,
I = $\frac{2}{3}\int_{0}^{\pi}cos^2tdt$
I = $\frac{1}{3}\int_{0}^{\pi}(1+cos2t)dt$
I = $\frac{1}{3}\left[t+\frac{sin2t}{t}\right]_{0}^{\pi}$
I = $\frac{1}{3}\left[\pi+0-0-0\right]$
I = $\frac{\pi}{3}$
Therefore, the value of $\int_{0}^{\pi^\frac{3}{2}}\sqrt{x}cos^2x^\frac{2}{3}dx$ is $\frac{\pi}{3}$ .

Question 45. $\int_{1}^{2}\frac{1}{x(1+logx)^2}dx$

Solution:

We have,
I = $\int_{1}^{2}\frac{1}{x(1+logx)^2}dx$
Let 1 + log x = t. So, we have
=> 1/x dx = dt
Now, the lower limit is, x = 0
=> t = 1 + log x
=> t = 1 + log 0
=> t = 1
Also, the upper limit is, x = 2
=> t = 1 + log x
=> t = 1 + log 2
So, the equation becomes,
I = $\int_{1}^{1+\log2}\frac{1}{t^2}dt$
I = $\left[\frac{-1}{t}\right]_{1}^{1+\log2}$
I = $\frac{-1}{1+\log2}+1$
I = $\frac{\log2}{1+\log2}$
Therefore, the value of $\int_{1}^{2}\frac{1}{x(1+logx)^2}dx$ is $\frac{\log2}{1+\log2}$ .

Question 46. $\int_{0}^{\frac{\pi}{2}}\cos^5xdx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}\cos^5xdx$
I = $\int_{0}^{\frac{\pi}{2}}(1-\sin^2x)^2\cos xdx$
Let sin x = t. So, we have
=> cos x dx = dt
Now, the lower limit is, x = 0
=> t = sin x
=> t = sin 0
=> t = 0
Also, the upper limit is, x = π/2
=> t = sin x
=> t = sin π/2
=> t = 1
So, the equation becomes,
I = $\int_{0}^{1}(1-t^2)^2dt$
I = $\int_{0}^{1}(1+t^4-2t^2)dt$
I = $\left[t-\frac{2}{3}t^3+\frac{t^5}{5}\right]_{0}^{1}$
I = $1-\frac{2}{3}+\frac{1}{5}$
I = $\frac{8}{15}$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\cos^5xdx$ is $\frac{8}{15}$ .

Question 47. $\int_{4}^{9}\frac{\sqrt{x}}{(30-x^{\frac{3}{2}})^2}dx$

Solution:

We have,
I = $\int_{4}^{9}\frac{\sqrt{x}}{(30-x^{\frac{3}{2}})^2}dx$
Let 30 – x^3/2 = t. So, we have
=> $-\frac{3}{2}\sqrt{x}dx$ = dt
=> $\frac{3}{2}\sqrt{x}dx$ = – dt
Now, the lower limit is, x = 4
=> t = 30 – x^3/2
=> t = 30 – 4^3/2
=> t = 30 – 8
=> t = 22
Also, the upper limit is, x = 9
=> t = 30 – x^3/2
=> t = 30 – 9^3/2
=> t = 30 – 27
=> t = 3
So, the equation becomes,
I = $\int_{22}^{3}\frac{-2}{3t^2}dt$
I = $\frac{2}{3}\int_{3}^{22}\frac{1}{t^2}dt$
I = $\frac{2}{3}\left[\frac{-1}{t}\right]_{3}^{22}$
I = $\frac{2}{3}\left[\frac{-1}{22}+\frac{1}{3}\right]$
I = $\frac{2}{3}(\frac{19}{66})$
I = $\frac{19}{99}$
Therefore, the value of $\int_{4}^{9}\frac{\sqrt{x}}{(30-x^{\frac{3}{2}})^2}dx$ is $\frac{19}{99}$ .

Question 48. $\int_{0}^{\pi}sin^3x(1+2cosx)(1+cosx)^2dx$

Solution:

We have,
I = $\int_{0}^{\pi}sin^3x(1+2cosx)(1+cosx)^2dx$
Let cos x = t. So, we have
=> – sin x dx = dt
=> sin x dx = –dt
Now, the lower limit is, x = 0
=> t = cos x
=> t = cos 0
=> t = 1
Also, the upper limit is, x = π
=> t = cos x
=> t = cos π
=> t = –1
So, the equation becomes,
I = $\int_{0}^{\pi}sin^2x(1+2cosx)(1+cosx)^2.sinxdx$
I = $\int_{1}^{-1}-(1-t^2)(1+2t)(1+t)^2dt$
I = $\int_{-1}^{1}(1+2t-t^2-2t^3)(1+t^2+2t)dt$
I = $\int_{-1}^{1}(1+4t+4t^2-2t^3-5t^4-2t^5)dt$
I = $\left[t+2t^2+\frac{4}{3}t^3-\frac{1}{2}t^4-t^5-\frac{1}{3}t^6\right]_{-1}^{1}$
I = $2+0+\frac{8}{3}-0-2-0$
I = $\frac{8}{3}$
Therefore, the value of $\int_{0}^{\pi}sin^3x(1+2cosx)(1+cosx)^2dx$ is $\frac{8}{3}$ .

Question 49. $\int_{0}^{\frac{\pi}{2}}2sinxcosxtan^{-1}(sinx)dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}2sinxcosxtan^{-1}(sinx)dx$
Let sin x = t. So, we have
=> cos x dx = dt
Now, the lower limit is, x = 0
=> t = sin x
=> t = sin 0
=> t = 0
Also, the upper limit is, x = π/2
=> t = sin x
=> t = sin π/2
=> t = 1
So, the equation becomes,
I = $2\int_{0}^{1}ttan^{-1}tdt$
I = $2\left[\frac{1}{2}t^2tan^{-1}t-\frac{t}{2}+\frac{1}{2}tan^{-1}t\right]_0^1$
I = $2(\frac{\pi}{4}-\frac{1}{2})$
I = $\frac{\pi}{2}-1$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}2sinxcosxtan^{-1}(sinx)dx$ is $\frac{\pi}{2}-1$ .

Question 50. $\int_{0}^{\frac{\pi}{2}}sin2xtan^{-1}(sinx)dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}sin2xtan^{-1}(sinx)dx$
I = $\int_{0}^{\frac{\pi}{2}}2sinxcosxtan^{-1}(sinx)dx$
Let sin x = t. So, we have
=> cos x dx = dt
Now, the lower limit is, x = 0
=> t = sin x
=> t = sin 0
=> t = 0
Also, the upper limit is, x = π/2
=> t = sin x
=> t = sin π/2
=> t = 1
So, the equation becomes,
I = $2\int_{0}^{1}ttan^{-1}tdt$
I = $2\left[\frac{1}{2}t^2tan^{-1}t-\frac{t}{2}+\frac{1}{2}tan^{-1}t\right]_0^1$
I = $2(\frac{\pi}{4}-\frac{1}{2})$
I = $\frac{\pi}{2}-1$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}sin2xtan^{-1}(sinx)dx$ is $\frac{\pi}{2}-1$ .

Question 51. $\int_{0}^{1}(cos^{-1}x)^2dx$

Solution:

We have,
I = $\int_{0}^{1}(cos^{-1}x)^2dx$
On using integration by parts, we get,
I = $(cos^{-1}x)^2\int_{0}^{1}dx-\int_0^1(\int dx)\frac{d}{dx}(cos^{-1}x)^2dx$
I = $\left[x(cos^{-1}x)^2\right]_{0}^{1}+2\int_0^1\frac{xcos^{-1}x}{\sqrt{1-x^2}}dx$
Let cos^-1 x = t. So, we have
=> $\frac{-1}{\sqrt{1-x^2}}dx$ = dt
Now, the lower limit is, x = 0
=> t = cos^-1 x
=> t = cos^-1 0
=> t = π/2
Also, the upper limit is, x = 1
=> t = cos^-1 x
=> t = cos^-1 1
=> t = 0
So, the equation becomes,
I = $\left[x(cos^{-1}x)^2\right]_{0}^{1}+2\int_\frac{\pi}{2}^0-tcostdt$
I = $0-0+2\int^\frac{\pi}{2}_0tcostdt$
I = $2\int^\frac{\pi}{2}_0tcostdt$
I = $t\int_{0}^{\frac{\pi}{2}}costdt-\int_0^\frac{\pi}{2}(\int costdt)\frac{dt}{dt}dt$
I = $2\left[tsint-\int sintdt\right]^\frac{\pi}{2}_0$
I = $2\left[tsint+cost\right]^\frac{\pi}{2}_0$
I = $2(\frac{\pi}{2}-1)$
I = π – 2
Therefore, the value of $\int_{0}^{1}(cos^{-1}x)^2dx$ is π – 2.

Question 52. $\int_{0}^{a}sin^{-1}\sqrt{\frac{x}{a+x}}dx$

Solution:

We have,
I = $\int_{0}^{a}sin^{-1}\sqrt{\frac{x}{a+x}}dx$
Let x = a tan² t. So, we have
=> dx = 2a tan t sec²t dt
Now, the lower limit is, x = 0
=> a tan² t = x
=> a tan² t = 0
=> tan t = 0
=> t = 0
Also, the upper limit is, x = a
=> a tan² t = x
=> a tan² t = a
=> tan² t = 1
=> tan t = 1
=> t = π/4
So, the equation becomes,
I = $\int_{0}^{\frac{\pi}{4}}sin^{-1}\sqrt{\frac{atan^2t}{a+atan^2t}}(2atantsec^2t)dt$
I = $\int_{0}^{\frac{\pi}{4}}sin^{-1}\sqrt{\frac{atan^2t}{a(1+tan^2t)}}(2atantsec^2t)dt$
I = $\int_{0}^{\frac{\pi}{4}}sin^{-1}\sqrt{\frac{tan^2t}{sec^2t}}(2atantsec^2t)dt$
I = $2a\int_{0}^{\frac{\pi}{4}}sin^{-1}(sint)(tantsec^2t)dt$
I = $2a\int_{0}^{\frac{\pi}{4}}t(tantsec^2t)dt$
I = $2a\left[\frac{ttan^2t}{2}-\int \frac{tan^2t}{2}dt\right]_0^\frac{\pi}{4}$
I = $\left[attan^2t-\frac{2a}{2}\int (sec^2t-1)dt\right]_0^\frac{\pi}{4}$
I = $\left[attan^2t-atant+at\right]_0^\frac{\pi}{4}$
I = $\frac{a\pi}{4}tan^2\frac{\pi}{4}-atan\frac{\pi}{4}+a\frac{\pi}{4}$
I = $\frac{a\pi}{4}-a+\frac{a\pi}{4}$
I = $\frac{a\pi}{2}-a$
I = $a(\frac{\pi}{2}-1)$
Therefore, the value of $\int_{0}^{a}sin^{-1}\sqrt{\frac{x}{a+x}}dx$ is $a(\frac{\pi}{2}-1)$ .

Question 53. $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sqrt{1+cosx}}{(1-cosx)^{\frac{3}{2}}}dx$

Solution:

We have,
I = $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sqrt{1+cosx}}{(1-cosx)^{\frac{3}{2}}}dx$
I = $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sqrt{2cos^2\frac{x}{2}}}{(2sin^2\frac{x}{2})^{\frac{3}{2}}}dx$
I = $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sqrt{2}cos\frac{x}{2}}{2\sqrt{2}sin^3\frac{x}{2}}dx$
I = $\frac{1}{2}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{cos\frac{x}{2}}{sin^3\frac{x}{2}}dx$
I = $\frac{1}{2}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}cot\frac{x}{2}\cosec^2\frac{x}{2}dx$
Let cot x/2 = t. So, we have
=> $\frac{-1}{2}cosec^2\frac{x}{2}dx$ = dt
Now, the lower limit is, x = π/3
=> t = cot x/2
=> t = cot π/6
=> t = √3
Also, the upper limit is, x = π/2
=> t = cot x/2
=> t = cot π/4
=> t = 1
So, the equation becomes,
I = $-\int_{\sqrt{3}}^{1}tdt$
I = $\int^{\sqrt{3}}_{1}tdt$
I = $\left[\frac{t^2}{2}\right]^{\sqrt{3}}_{1}$
I = $\frac{3}{2}-\frac{1}{2}$
I = 1
Therefore, the value of $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sqrt{1+cosx}}{(1-cosx)^{\frac{3}{2}}}dx$ is 1.

Question 54. $\int_{0}^{a}x\sqrt{\frac{a^2-x^2}{a^2+x^2}}dx$

Solution:

We have,
I = $\int_{0}^{a}x\sqrt{\frac{a^2-x^2}{a^2+x^2}}dx$
Let x² = a² cos 2t. So, we have
=> 2x dx = – 2a² sin 2t dt
Now, the lower limit is, x = 0
=> a² cos 2t = x²
=> a² cos 2t = 0
=> cos 2t = 0
=> 2t = π/2
=> t = π/4
Also, the upper limit is, x = a
=> a² cos 2t = x²
=> a² cos 2t = a²
=> cos 2t = 1
=> 2t = 0
=> t = 0
So, the equation becomes,
I = $\int_{\frac{\pi}{4}}^{0}\sqrt{\frac{a^2-a^2cos2t}{a^2-(1-cos2t)}}(-a^2sin2t)dt$
I = $-a^2\int_{\frac{\pi}{4}}^{0}\frac{sint}{cost}sin2tdt$
I = $a^2\int^{\frac{\pi}{4}}_{0}\frac{sint}{cost}sin2tdt$
I = $a^2\int^{\frac{\pi}{4}}_{0}2sin^2tdt$
I = $a^2\int^{\frac{\pi}{4}}_{0}(1-cos2t)dt$
I = $a^2\left[t-\frac{sin2t}{2}\right]^{\frac{\pi}{4}}_{0}$
I = $a^2(\frac{\pi}{4}-\frac{1}{2})$
Therefore, the value of $\int_{0}^{a}x\sqrt{\frac{a^2-x^2}{a^2+x^2}}dx$ is $a^2(\frac{\pi}{4}-\frac{1}{2})$ .

Question 55. $\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}dx$

Solution:

We have,
I = $\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}dx$
Let x = a cos 2t. So, we have
=> dx = –2a sin 2t
Now, the lower limit is, x = –a
=> a cos 2t = x
=> a cos 2t = –a
=> cos 2t = –1
=> 2t = π
=> t = π/2
Also, the upper limit is, x = a
=> a cos 2t = x
=> a cos 2t = a
=> cos 2t = 1
=> 2t = 0
=> t = 0
So, the equation becomes,
I = $\int_{\frac{\pi}{2}}^{0}\sqrt{\frac{a-acos2t}{a+acos2t}}(-2asin2t)dt$
I = $-2a\int_{\frac{\pi}{2}}^{0}\sqrt{\frac{a(1-cos2t)}{a(1+cos2t)}}sin2tdt$
I = $-2a\int_{\frac{\pi}{2}}^{0}\sqrt{\frac{1-cos2t}{1+cos2t}}sin2tdt$
I = $-2a\int_{\frac{\pi}{2}}^{0}\sqrt{\frac{2sin^2t}{2cos^2t}}sin2tdt$
I = $-2a\int_{\frac{\pi}{2}}^{0}\frac{sint}{cost}(2sintcost)dt$
I = $-4a\int_{\frac{\pi}{2}}^{0}sin^2tdt$
I = $\frac{4a}{2}\int^{\frac{\pi}{2}}_{0}(1-cos2t)dt$
I = $2a\int^{\frac{\pi}{2}}_{0}(1-cos2t)dt$
I = $2a\left[t-\frac{sin2t}{2}\right]^{\frac{\pi}{2}}_{0}$
I = $2a\left[\frac{\pi}{2}-0-0+0\right]$
I = πa
Therefore, the value of $\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}dx$ is πa.

Question 56. $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{cos^2x+3cosx+2}dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{cos^2x+3cosx+2}dx$
Let cos x = t. So, we have
=> – sin x dx = dt
=> sin x dx = –dt
Now, the lower limit is, x = 0
=> t = cos x
=> t = cos 0
=> t = 1
Also, the upper limit is, x = π/2
=> t = cos x
=> t = cos π/2
=> t = 0
So, the equation becomes,
I = $\int_{1}^{0}\frac{-t}{t^2+3t+2}dt$
I = $\int_{0}^{1}\frac{t}{(t+2)(t+1)}dt$
I = $\int_{0}^{1}(\frac{-1}{t+1}+\frac{2}{t+2})dt$
I = $\left[-log|1+t|+2log|t+2|\right]_{0}^{1}$
I = – log 2 + 2 log 3 + 0 – 2 log 2
I = 2 log 3 – 3 log 2
I = log 9 – log 8
I = log 9/8
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{cos^2x+3cosx+2}dx$ is log 9/8.

Question 57. $\int_{0}^{\frac{\pi}{2}}\frac{tanx}{1+m^2tan^2x}dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}\frac{tanx}{1+m^2tan^2x}dx$
I = $\int_{0}^{\frac{\pi}{2}}\frac{\frac{sinx}{cosx}}{1+m^2(\frac{sin^2x}{cos^2x})}dx$
I = $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{cos^2x+m^2sin^2x}dx$
Let sin² x = t. So, we have
=> 2 sin x cos x dx = dt
Now, the lower limit is, x = 0
=> t = sin² x
=> t = sin² 0
=> t = 0
Also, the upper limit is, x = π/2
=> t = sin² x
=> t = sin² π/2
=> t = 1
So, the equation becomes,
I = $\frac{1}{2}\int_{0}^{1}\frac{1}{(1-t)+m^2t}dt$
I = $\frac{1}{2}\int_{0}^{1}\frac{1}{(m^2-1)t+1}dt$
I = $\frac{1}{2(m^2-1)}\left[\log|(m^2-1)t+1|\right]_0^1$
I = $\frac{1}{2(m^2-1)}\left[\log|m^2-1+1|-log1\right]$
I = $\frac{1}{2(m^2-1)}\left[\log m^2-log1\right]$
I = $\frac{\log m^2}{2(m^2-1)}$
I = $\frac{2\log m}{2(m^2-1)}$
I = $\frac{\log m}{m^2-1}$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{tanx}{1+m^2tan^2x}dx$ is $\frac{\log m}{m^2-1}$ .

Question 58. $\int_{0}^{\frac{1}{2}}\frac{1}{(1+x^2)(\sqrt{1-x^2})}dx$

Solution:

We have,
I = $\int_{0}^{\frac{1}{2}}\frac{1}{(1+x^2)(\sqrt{1-x^2})}dx$
Let x = sin t. So, we have
=> dx = cos t dt
Now, the lower limit is, x = 0
=> sin t = x
=> sin t = 0
=> t = 0
Also, the upper limit is, x = 1/2
=> sin t = x
=> sin t = 1/2
=> t = π/6
So, the equation becomes,
I = $\int_{0}^{\frac{\pi}{6}}\frac{1}{(1+sin^2t)(\sqrt{1-sin^2t})}(cost)dt$
I = $\int_{0}^{\frac{\pi}{6}}\frac{1}{(1+sin^2t)(\sqrt{cos^2t})}(cost)dt$
I = $\int_{0}^{\frac{\pi}{6}}\frac{1}{(1+sin^2t)cost}(cost)dt$
I = $\int_{0}^{\frac{\pi}{6}}\frac{1}{1+sin^2t}dt$
I = $\int_{0}^{\frac{\pi}{6}}\frac{sec^2t}{1+2tan^2t}dt$
Let tan t = u. So, we have
=> sec² t dt = du
Now, the lower limit is, t = 0
=> u = tan t
=> u = tan 0
=> t = 0
Also, the upper limit is, t = π/6
=> u = tan t
=> u = tan π/6
=> t = 1/√3
So, the equation becomes,
I = $\int_{0}^{\frac{1}{\sqrt{3}}}\frac{1}{1+2u^2}du$
I = $\frac{1}{\sqrt{2}}\left[tan^{-1}(\sqrt{2}u)\right]_{0}^{\frac{1}{\sqrt{3}}}$
I = $\frac{1}{\sqrt{2}}\tan^{-1}\sqrt{\frac{2}{3}}$
Therefore, the value of $\int_{0}^{\frac{1}{2}}\frac{1}{(1+x^2)(\sqrt{1-x^2})}dx$ is $\frac{1}{\sqrt{2}}\tan^{-1}\sqrt{\frac{2}{3}}$ .

Question 59. $\int_{\frac{1}{3}}^{1}\frac{(x-x^3)^{\frac{1}{3}}}{x^4}dx$

Solution:

We have,
I = $\int_{\frac{1}{3}}^{1}\frac{(x-x^3)^{\frac{1}{3}}}{x^4}dx$
Let 1/x² – 1 = t. So, we have
=> –2/x³ dx = dt
Now, the lower limit is, x = 1/3
=> t = 1/x² – 1
=> t = 9 – 1
=> t = 8
Also, the upper limit is, x = 1
=> t = 1/x² – 1
=> t = 1 – 1
=> t = 0
So, the equation becomes,
I = $\frac{-1}{2}\int_{8}^{0}t^{\frac{1}{3}}dt$
I = $\frac{1}{2}\left[\frac{t^{\frac{4}{3}}}{\frac{4}{3}}\right]_{0}^{8}$
I = $\frac{3}{8}\left[t^{\frac{4}{3}}\right]_{0}^{8}$
I = $\frac{3}{8}(8^{\frac{4}{3}})$
I = $\frac{3}{8}(16)$
I = 6
Therefore, the value of $\int_{\frac{1}{3}}^{1}\frac{(x-x^3)^{\frac{1}{3}}}{x^4}dx$ is 6.

Question 60. $\int_{0}^{\frac{\pi}{4}}\frac{sin^2xcos^2x}{(sin^3x+cos^3x)^2}dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{4}}\frac{sin^2xcos^2x}{(sin^3x+cos^3x)^2}dx$
I = $\int_{0}^{\frac{\pi}{4}}\frac{tan^2xsec^2x}{tan^6x+2tan^3x+1}dx$
Let tan x = t. So, we have
=> sec² x dx = dt
Now, the lower limit is, x = 0
=> t = tan t
=> t = tan 0
=> t = 0
Also, the upper limit is, x = π/4
=> t = tan t
=> t = tan π/4
=> t = 1
So, the equation becomes,
I = $\int_{0}^{1}\frac{t^2}{t^6+2t^3+1}dt$
Let t³= u. So, we have
=> 3t² dt = du
=> t² dt = du/3
Now, the lower limit is, t = 0
=> u = t³
=> u = 0³
=> u = 0
Also, the upper limit is, t = 1
=> u = t³
=> u = 1³
=> u = 1
So, the equation becomes,
I = $\frac{1}{3}\int_{0}^{1}\frac{1}{u^2+2u+1}du$
I = $\frac{1}{3}\int_{0}^{1}\frac{1}{(u+1)^2}du$
I = $\frac{1}{3}\left[\frac{-1}{u+1}\right]_{0}^{1}$
I = $\frac{1}{3}\left[\frac{-1}{1+1}+\frac{1}{1+0}\right]$
I = $\frac{1}{3}(\frac{-1}{2}+1)$
I = $\frac{1}{3}(\frac{1}{2})$
I = $\frac{1}{6}$
Therefore, the value of $\int_{0}^{\frac{\pi}{4}}\frac{sin^2xcos^2x}{(sin^3x+cos^3x)^2}dx$ is $\frac{1}{6}$ .

Question 61. $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx-cos^3x}(sec^2x-1)cos^2xdx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx-cos^3x}(sec^2x-1)cos^2xdx$
I = $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx-cos^3x}tan^2xcos^2xdx$
I = $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx-cos^3x}sin^2xdx$
I = $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx(1-cos^2x)}sin^2xdx$
I = $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx}sin^3xdx$
Let cos x = t. So, we have
=> – sin x dx = dt
Now, the lower limit is, x = 0
=> t = cos x
=> t = cos 0
=> t = 1
Also, the upper limit is, x = π/2
=> t = cos x
=> t = cos π/2
=> t = 0
So, the equation becomes,
I = $-\int_{1}^{0}\sqrt{t}(1-t^2)dt$
I = $\int_{0}^{1}\sqrt{t}(1-t^2)dt$
I = $\int_{0}^{1}(t^{\frac{1}{2}}-t^{\frac{5}{2}})dt$
I = $\left[\frac{2t^{\frac{3}{2}}}{3}-\frac{2t^{\frac{7}{2}}}{7}\right]_{0}^{1}$
I = $\frac{2}{3}-\frac{2}{7}$
I = $\frac{8}{21}$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx-cos^3x}(sec^2x-1)cos^2xdx$ is $\frac{8}{21}$ .

Question 62. $\int_{0}^{\frac{\pi}{2}}\frac{cosx}{(cos\frac{x}{2}+sin\frac{x}{2})^n}dx$

Solution:

We have,
I = $\int_{0}^{\frac{\pi}{2}}\frac{cosx}{(cos\frac{x}{2}+sin\frac{x}{2})^n}dx$
I = $\int_{0}^{\frac{\pi}{2}}\frac{cos^2\frac{x}{2}-sin^2\frac{x}{2}}{(cos\frac{x}{2}+sin\frac{x}{2})^n}dx$
I = $\int_{0}^{\frac{\pi}{2}}\frac{cos\frac{x}{2}-sin\frac{x}{2}}{(cos\frac{x}{2}+sin\frac{x}{2})^{n-1}}dx$
Let cos x/2 + sin x/2 = t. So, we have
=> (cos x/2 – sin x/2) dx = 2 dt
Now, the lower limit is, x = 0
=> t = cos x/2 + sin x/2
=> t = cos 0 + sin 0
=> t = 1 + 0
=> t = 1
Also, the upper limit is, x = π/2
=> t = cos x/2 + sin x/2
=> t = cos π/2 + sin π/2
=> t = 1/√2 + 1/√2
=> t = √2
So, the equation becomes,
I = $\int_{1}^{\sqrt{2}}\frac{2}{(t)^{n-1}}dt$
I = $\left[\frac{2t^{-n+2}}{-n+2}\right]_{1}^{\sqrt{2}}$
I = $\frac{2}{2-n}\left[(\sqrt{2})^{2-n}-1\right]$
I = $\frac{2}{2-n}\left[2^{1-\frac{n}{2}}-1\right]$
Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{cosx}{(cos\frac{x}{2}+sin\frac{x}{2})^n}dx$ is $\frac{2}{2-n}\left[2^{1-\frac{n}{2}}-1\right]$ .

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RD Sharma Class 12 Ex 20.2 Solutions Chapter 20 Definite Integrals

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

Evaluate the following definite integrals:

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.

Question 14.

Question 15.

Question 16.

Question 17.

Question 18.

Question 19.

Question 20.

Evaluate the following definite integrals:

Question 21.

Question 22.

Question 23.

Question 24.

Question 25.

Question 26.

Question 27.

Question 28.

Question 29.

Question 30.

Question 31.

Question 32.

Question 33.

Question 34.

Question 35.

Question 36.

Question 37.

Question 38.

Question 39.

Question 40.

Question 41.

Evaluate the following definite integrals:

Question 42.

Question 43.

Question 44.

Question 45.

Question 46.

Question 47.

Question 48.

Question 49.

Question 50.

Question 51.

Question 52.

Question 53.

Question 54.

Question 55.

Question 56.

Question 57.

Question 58.

Question 59.

Question 60.

Question 61.

Question 62.

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Question 1. $\int_{2}^{4}\frac{x}{x^2+1}dx$

Question 2. $\int_{1}^{2}\frac{1}{x(1+logx)^2}dx$

Question 3. $\int_{1}^{2}\frac{3x}{9x^2-1}dx$

Question 4. $\int_{0}^{\frac{\pi}{2}}\frac{1}{5cosx+3sinx}dx$

Question 5. $\int_{0}^{a}\frac{x}{\sqrt{a^2+x^2}}dx$

Question 6. $\int_{0}^{1}\frac{e^x}{1+e^{2x}}dx$

Question 7. $\int_{0}^{1}xe^{x^2}dx$

Question 8. $\int_{1}^{3}\frac{cos(logx)}{x}dx$

Question 9. $\int_{0}^{1}\frac{2x}{1+x^4}dx$

Question 10. $\int_{0}^{a}\sqrt{a^2-x^2}dx$

Question 11. $\int_{0}^{\frac{\pi}{2}}\sqrt{sinØ}cos^5ØdØ$

Question 12. $\int_{0}^{\frac{\pi}{2}}\frac{cosx}{1+sinx}dx$

Question 13. $\int_{0}^{\frac{\pi}{2}}\frac{sin\theta}{\sqrt{1+cos\theta}}d\theta$

Question 14. $\int_{0}^{\frac{\pi}{3}}\frac{cosx}{3+4sinx}dx$

Question 15. $\int_{0}^{1}\frac{\sqrt{tan^{-1}x}}{1+x^2}dx$

Question 16. $\int_{0}^{2}x\sqrt{x+2}dx$

Question 17. $\int_{0}^{1}tan^{-1}\frac{2x}{1-x^2}dx$

Question 18. $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{1+sin^4x}dx$

Question 19. $\int_{0}^{\frac{\pi}{2}}\frac{1}{acosx+bsinx}dx$

Question 20. $\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4sinx}dx$

Question 21. $\int_{0}^{\pi}\frac{sinx}{sinx+cosx}dx$

Question 22. $\int_{0}^{\pi}\frac{1}{3+2sinx+cosx}dx$

Question 23. $\int_{0}^{1}tan^{-1}xdx$

Question 24. $\int_{0}^{\frac{1}{2}}\frac{xsin^{-1}x}{\sqrt{1-x^2}}dx$

Question 25. $\int_{0}^{\frac{\pi}{4}}(\sqrt{tanx}+\sqrt{cotx})dx$

Question 26. $\int_{0}^{\frac{\pi}{4}}\frac{tan^3x}{1+cos2x}dx$

Question 27. $\int_{0}^{\pi}\frac{1}{5+3cosx}dx$

Question 28. $\int_{0}^{\frac{\pi}{2}}\frac{1}{a^2sin^2x+b^2cos^2x}dx$

Question 29. $\int_{0}^{\frac{\pi}{2}}\frac{x+sinx}{1+cosx}dx$

Question 30. $\int_{0}^{1}\frac{tan^{-1}x}{1+x^2}dx$

Question 31. $\int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{3+sin2x}dx$

Question 32. $\int_{0}^{1}xtan^{-1}xdx$

Question 33. $\int_{0}^{1}\frac{1-x^2}{x^4+x^2+1}dx$

Question 34. $\int_{0}^{1}\frac{24x^3}{(1+x^2)^4}dx$

Question 35. $\int_{4}^{12}x(x-4)^{\frac{1}{3}}dx$

Question 36. $\int_{0}^{\frac{\pi}{2}}x^2sinxdx$

Question 37. $\int_{0}^{1}\sqrt{\frac{1-x}{1+x}}dx$

Question 38. $\int_{0}^{1}\frac{1-x^2}{(1+x^2)^2}dx$

Question 39. $\int_{-1}^{1}5x^4\sqrt{x^2+1}dx$

Question 40. $\int_{0}^{\frac{\pi}{2}}\frac{cos^2x}{1+3sin^2x}dx$

Question 41. $\int_{0}^{\frac{\pi}{4}}sin^32tcos2tdt$

Question 42. $\int_{0}^{\pi}5(5-4\cos\theta)^{\frac{1}{4}}\sin\theta d\theta$

Question 43. $\int_{0}^{\frac{\pi}{6}}\cos^{-3}2\theta \sin2\theta d\theta$

Question 44. $\int_{0}^{\pi^\frac{3}{2}}\sqrt{x}cos^2x^\frac{2}{3}dx$

Question 45. $\int_{1}^{2}\frac{1}{x(1+logx)^2}dx$

Question 46. $\int_{0}^{\frac{\pi}{2}}\cos^5xdx$

Question 47. $\int_{4}^{9}\frac{\sqrt{x}}{(30-x^{\frac{3}{2}})^2}dx$

Question 48. $\int_{0}^{\pi}sin^3x(1+2cosx)(1+cosx)^2dx$

Question 49. $\int_{0}^{\frac{\pi}{2}}2sinxcosxtan^{-1}(sinx)dx$

Question 50. $\int_{0}^{\frac{\pi}{2}}sin2xtan^{-1}(sinx)dx$

Question 51. $\int_{0}^{1}(cos^{-1}x)^2dx$

Question 52. $\int_{0}^{a}sin^{-1}\sqrt{\frac{x}{a+x}}dx$

Question 53. $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sqrt{1+cosx}}{(1-cosx)^{\frac{3}{2}}}dx$

Question 54. $\int_{0}^{a}x\sqrt{\frac{a^2-x^2}{a^2+x^2}}dx$

Question 55. $\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}dx$

Question 56. $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{cos^2x+3cosx+2}dx$

Question 57. $\int_{0}^{\frac{\pi}{2}}\frac{tanx}{1+m^2tan^2x}dx$

Question 58. $\int_{0}^{\frac{1}{2}}\frac{1}{(1+x^2)(\sqrt{1-x^2})}dx$

Question 59. $\int_{\frac{1}{3}}^{1}\frac{(x-x^3)^{\frac{1}{3}}}{x^4}dx$

Question 60. $\int_{0}^{\frac{\pi}{4}}\frac{sin^2xcos^2x}{(sin^3x+cos^3x)^2}dx$

Question 61. $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx-cos^3x}(sec^2x-1)cos^2xdx$

Question 62. $\int_{0}^{\frac{\pi}{2}}\frac{cosx}{(cos\frac{x}{2}+sin\frac{x}{2})^n}dx$