RD Sharma Class 12 Ex 20.2 Solutions Chapter 20 Definite Integrals

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter20
Exercise20.2
CategoryRD Sharma Solutions

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 = a+ x2

=> t2 = a+ 02

=> t2 = a2

=> t = a

Also, the upper limit is, x = a

=> t2 = a2 + x2

=> t= 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 ex = t. So, we have

=> ex dx = dt

Now, the lower limit is, x = 0

=> t = ex

=> t = e0

=> t = 1

Also, the upper limit is, x = a

=> t = ex

=> 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

=> t= 2

=> t = √2

Also, the upper limit is, θ = π/2

=> t2 = 1 + cos θ

=> t= 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–1 1

=> 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

=> t= 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

=> secx/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

=> secx 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

=> secx 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 – x3/2 = 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 – x3/2

=> t = 30 – 43/2

=> t = 30 – 8

=> t = 22

Also, the upper limit is, x = 9

=> t = 30 – x3/2

=> t = 30 – 93/2

=> 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 sect 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 t= 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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