RD Sharma Class 12 Ex 20.1 Solutions Chapter 20 Definite Integrals

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter20
Exercise20.1
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 20.1 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_{4}^{9} \frac{1}{\sqrt{x}}dx

Solution:

We have,

I = \int_{4}^{9} \frac{1}{\sqrt{x}}dx

I = \left[\frac{x^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\right]^9_4

I = \left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]^9_4

I = \left[2\sqrt{x}\right]^9_4

I = 2[√9 – √4 ] 

I = 2 (3 − 2)

I = 2 (1)

I = 2

Therefore, the value of \int_{4}^{9} \frac{1}{\sqrt{x}}dx       is 2.

Question 2. \int_{-2}^{3} \frac{1}{x+7}dx

Solution:

We have,

I = \int_{-2}^{3} \frac{1}{x+7}dx

I = \left[log(x+7)\right]^3_{-2}

I = log (3 + 7) − log (−2 + 7)

I = log 10 − log 5

I = log\frac{10}{5}

I = log 2

Therefore, the value of \int_{-2}^{3} \frac{1}{x+7}dx       is log 2.

Question 3. \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1-x^2}}dx

Solution:

We have,

I = \int_{0}^{\frac{1}{2}} \frac{1}{\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 = 0

=> t = 0

Also, the upper limit is,

=> x = 1/2

=> sin t = 1/2

=> t = π/6

So, the equation becomes,

I = \int_{0}^{\frac{\pi}{6}} \frac{1}{\sqrt{1-sin^2t}}costdt

I = \int_{0}^{\frac{\pi}{6}} \frac{1}{\sqrt{cos^2t}}costdt

I = \int_{0}^{\frac{\pi}{6}} (\frac{1}{cost})costdt

I = \int_{0}^{\frac{\pi}{6}} 1dt

I = \left[t\right]_0^{\frac{\pi}{6}}

I =  π/6 – 0

I = π/6​

Therefore, the value of \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1-x^2}}dx       is π/6.

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

Solution:

We have,

I = \int_{0}^{1} \frac{1}{1+x^2}dx

I = \left[tan^{-1}x\right]_0^1

I = tan^{-1}1-tan^{-1}0

I = \frac{\pi}{4}-0

I = π/4

Therefore, the value of \int_{0}^{1} \frac{1}{1+x^2}dx       is π/4.

Question 5. \int_{2}^{3} \frac{x}{x^2+1}dx

Solution:

We have,

I = \int_{2}^{3} \frac{x}{x^2+1}dx

Let x2 + 1 = t, so we have,

=> 2x dx = dt

=> x dx = dt/2

Now, the lower limit is, x = 2

=> t = x2 + 1

=> t = (2)2 + 1

=> t = 4 + 1

=> t = 5

Also, the upper limit is, x = 3

=> t = x2 + 1

=> t = (3)+ 1

=> t = 9 + 1

=> t = 10

So, the equation becomes,

I = \int_{5}^{10} \frac{1}{2t}dt

I = \frac{1}{2}\int_{5}^{10} \frac{1}{t}dt

I = \frac{1}{2}\left[logt\right]^{10}_5

I = 1/2[log10 – log5] 

I = 1/2[log10/5]

I = 1/2[log2]

I = log√2

Therefore, the value of \int_{2}^{3} \frac{x}{x^2+1}dx       is log√2.

Question 6. \int_{0}^{\infty} \frac{1}{a^2+b^2x^2}dx

Solution:

We have,

I = \int_{0}^{\infty} \frac{1}{a^2+b^2x^2}dx

I = \int_{0}^{\infty} \frac{1}{b^2}\left(\frac{1}{\frac{a^2}{b^2}+x^2}\right)dx

I = \frac{1}{b^2}\int_{0}^{\infty}\frac{1}{\frac{a^2}{b^2}+x^2}dx

I = \frac{1}{b^2}\left[\frac{b}{a}tan^{-1}\frac{bx}{a}\right]^{\infty}_{0}

I = \frac{1}{ab}\left[tan^{-1}\frac{bx}{a}\right]^{\infty}_{0}

I = 1/ab[tan-1∞ – tan-10] 

I = 1/ab[π/2 – 0]  

I = 1/ab[π/2]  

I = π/2ab   

Therefore, the value of \int_{0}^{\infty} \frac{1}{a^2+b^2x^2}dx       is π/2ab.

Question 7. \int_{-1}^{1} \frac{1}{1+x^2}dx

Solution:

We have,

I = \int_{-1}^{1} \frac{1}{1+x^2}dx

I = \left[tan^{-1}x\right]_{-1}^{1}

I = [tan-11 – tan-1(-1)]  

I = [π/4 – (-π/4)]   

I = [π/4 + π/4]  

I = 2π/4 

I = π/2  

Therefore, the value of \int_{-1}^{1} \frac{1}{1+x^2}dx       is π/2.

Question 8. \int_{0}^{\infty} e^{-x}dx

Solution:

We have,

I = \int_{0}^{\infty} e^{-x}dx

I = \left[-e^{-x}\right]^{\infty}_{0}

I = -e – (-e0

I = − 0 + 1

I = 1

Therefore, the value of \int_{0}^{\infty} e^{-x}dx       is 1.

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

Solution:

We have,

I = \int_{0}^{1} \frac{x}{x+1}dx

I = \int_{0}^{1} \frac{(x+1)-1}{x+1}dx

I = \int_{0}^{1} \frac{x+1}{x+1}dx-\int_{0}^{1}\frac{1}{x+1}dx

I = \int_{0}^{1} 1dx-\int_{0}^{1}\frac{1}{x+1}dx

I = \left[x\right]^1_0-\left[log(x+1)\right]^1_0

I = [1 − 0] − [log(1 + 1) − log(0 + 1)]

I = 1 − [log2 − log1]

I = 1 – log2/1 

I = 1 − log 2

I = log e − log 2

I = loge/2 

Therefore, the value of \int_{0}^{1} \frac{x}{x+1}dx        is loge/2.

Question 10. \int_{0}^{\frac{\pi}{2}} (sinx+cosx)dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} (sinx+cosx)dx

I = \int_{0}^{\frac{\pi}{2}} (sinx)dx+\int_{0}^{\frac{\pi}{2}}(cosx)dx

I = \left[-cosx\right]_{0}^{\frac{\pi}{2}}+\left[sinx\right]_{0}^{\frac{\pi}{2}}

I = [-cosπ/2 + cos0] + [sinπ/2 – sin0] 

I = [−0 + 1] + 1

I = 1 + 1

I = 2

Therefore, the value of \int_{0}^{\frac{\pi}{2}} (sinx+cosx)dx      is 2.

Question 11. \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} cotxdx

Solution:

We have,

I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} cotxdx

I = \left[log(sinx)\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}

I = log(sinπ/2) – log(sinπ/4)

I = log1 – log1/√2 

I = log\frac{1}{\frac{1}{\sqrt{2}}}

I = log√2  

Therefore, the value of \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} cotxdx     is log√2.

Question 12. \int_{0}^{\frac{\pi}{4}} secxdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{4}} secxdx

I = \left[log(secx+tanx)\right]^{\frac{\pi}{4}}_0

I = log(secπ/4 + tanπ/4 – log(sec0 + tan0) 

I = log(√2 + 1) – log(1 + 0) 

I = log(\frac{\sqrt{2}+1}{1})

I = log(√2 + 1) 

Therefore, the value of \int_{0}^{\frac{\pi}{4}} secxdx     is log(√2 + 1).

Question 13. \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} cosecxdx

Solution:

We have,

I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} cosecxdx

I = \left[log|cosecx-cotx|\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}}

I = [log|cosecπ/4 – cotπ/4|] – [log|cosecπ/6 – cotπ/6|]

I = [log|√2 – 1|] – [log|2 – √3|] 

I = log(\frac{\sqrt{2}-1}{2-\sqrt{3}})

Therefore, the value of \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} cosecxdx     is log(\frac{\sqrt{2}-1}{2-\sqrt{3}})     .

Question 14. \int_{0}^{1} \frac{1-x}{1+x}dx

Solution: 

We have, 

I = \int_{0}^{1} \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 = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is,

=> x = 1

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I = \int_{\frac{\pi}{4}}^{0} \frac{1-cos2t}{1+cos2t}(-2sin2t)dt

I = \int_{\frac{\pi}{4}}^{0}\frac{2sin^2t}{2cos^2t}(-2sin2t)dt

I = \int^{\frac{\pi}{4}}_{0}\frac{sin^2t}{cos^2t}(2sin2t)dt

I = \int^{\frac{\pi}{4}}_{0}\frac{sin^2t}{cos^2t}(4sintcost)dt

I = \int^{\frac{\pi}{4}}_{0}\frac{4sin^3t}{cost}dt

Let cos t = z, so we have,

=> – sin t dt = dz

=> sin t dt = – dz

Now, the lower limit is,

=> t = 0

=> z = cos t

=> z = cos 0

=> z = 1

Also, the upper limit is,

=> t = π/4

=> z = cos t

=> z = cos π/4

=> z = 1/√2

So, the equation becomes,

I = \int^{\frac{\pi}{4}}_{0}\frac{4sin^2t(sint)}{cost}dt

I = \int^{\frac{1}{\sqrt{2}}}_{1}\frac{-4(1-z^2)}{z}dz

I = -4\int^{\frac{1}{\sqrt{2}}}_{1}\frac{1-z^2}{z}dz

I = -4\int^{\frac{1}{\sqrt{2}}}_{1}(\frac{1}{z}-\frac{z^2}{z})dz

I = -4\int^{\frac{1}{\sqrt{2}}}_{1}(\frac{1}{z}-z)dz

I = -4\left[logz-\frac{z^2}{2}\right]^{\frac{1}{\sqrt{2}}}_1

I = -4[(log1/√2 – 1/2(2)) – (log1 – 1/2)]

I = -4[(log1/√2 – 1/4) – (0 – 1/2)]

I = -4[log1/√2 – 1/4 – 0 + 1/2]

I = -4[-log√2 + 1/4]

I = 4log√2 – 1 

I = 4 × 1/2log2 – 1

I = 2log2 – 1 

Therefore, the value of \int_{0}^{1} \frac{1-x}{1+x}dx     is 2log2 – 1.

Question 15. \int_{0}^{\pi} \frac{1}{1+sinx}dx

Solution: 

We have,

I = \int_{0}^{\pi} \frac{1}{1+sinx}dx

I = \int_{0}^{\pi} \frac{1-sinx}{(1+sinx)(1-sinx)}dx

I = \int_{0}^{\pi} \frac{1-sinx}{1-sin^2x}dx

I = \int_{0}^{\pi} \frac{1-sinx}{cos^2x}dx

I = \int_{0}^{\pi}\frac{1}{cos^2x}-\frac{sinx}{cos^2x}dx

I = \int_{0}^{\pi}sec^2x-\frac{sinx}{cosx(cosx)}dx

I = \int_{0}^{\pi}(sec^2x-tanxsecx)dx

I = \int_{0}^{\pi}sec^2xdx-\int_{0}^{\pi}tanxsecxdx

I = \left[tanx\right]^{\pi}_0-\left[secx\right]^{\pi}_0

I = [tan π – tan0] – [sec π – sec 0]

I = [0 – 0] – [–1 – 1]

I = 0 – (–2)

I = 2

Therefore, the value of \int_{0}^{\pi} \frac{1}{1+sinx}dx     is 2.

Question 16. \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1}{1+sinx}dx

Solution: 

We have,

I = \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1}{1+sinx}dx

I = \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1-sinx}{(1+sinx)(1-sinx)}dx

I = \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1-sinx}{1-sin^2x}dx

I = \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1-sinx}{cos^2x}dx

I = \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\frac{1}{cos^2x}-\frac{sinx}{cos^2x}dx

I = \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}sec^2x-\frac{sinx}{cosx(cosx)}dx

I = \int_{\frac{-\pi}{4}}^{\frac{-\pi}{4}}(sec^2x-tanxsecx)dx

I = \int_{\frac{-\pi}{4}}^{\frac{-\pi}{4}}sec^2xdx-\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}tanxsecxdx

I = \left[tanx\right]^{\frac{\pi}{4}}_{\frac{-\pi}{4}}-\left[secx\right]^{\frac{\pi}{4}}_{\frac{-\pi}{4}}

I = \left[tanx\right]^{\frac{\pi}{4}}_{\frac{-\pi}{4}}-\left[secx\right]^{\frac{\pi}{4}}_{\frac{-\pi}{4}}

I = [tan π/4 – tan(–π/4)] – [sec π/4 – sec (–π/4)]

I = [1 – (–1)] – [sec π/4 – sec (π/4)]

I = 2 – 0

I = 2

Therefore, the value of \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{1}{1+sinx}dx     is 2.

Question 17. \int_{0}^{\frac{\pi}{2}} cos^2xdx

Solution: 

We have,

I = \int_{0}^{\frac{\pi}{2}} cos^2xdx

I = \int_{0}^{\frac{\pi}{2}} (\frac{1+cos2x}{2})dx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} (1+cos2x)dx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}1dx+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}cos2xdx

I = \frac{1}{2}\left[x\right]^{\frac{\pi}{2}}_0+\frac{1}{2}\left[\frac{sin2x}{2}\right]^{\frac{\pi}{2}}_0

I = \frac{1}{2}\left[x\right]^{\frac{\pi}{2}}_0+\frac{1}{4}\left[sin2x\right]^{\frac{\pi}{2}}_0

I = 1/2[π/2 – 0] + 1/4[sinπ – sin0]

I = 1/2[π/2] + 1/4[0 – 0]

I = π/4 

Therefore, the value of \int_{0}^{\frac{\pi}{2}} cos^2xdx      is π/4.

Question 18. \int_{0}^{\frac{\pi}{2}} cos^3xdx

Solution: 

We have,

I = \int_{0}^{\frac{\pi}{2}} cos^3xdx

I = \int_{0}^{\frac{\pi}{2}} \frac{cos3x+3cosx}{4}dx

I = \frac{1}{4}\int_{0}^{\frac{\pi}{2}} (cos3x+3cosx)dx

I = \frac{1}{4}\int_{0}^{\frac{\pi}{2}} cos3xdx+\frac{3}{4}\int_{0}^{\frac{\pi}{2}}cosxdx

I = \frac{1}{4}\left[\frac{sin3x}{3}\right]^{\frac{\pi}{2}}_0+\frac{3}{4}[sinx]^{\frac{\pi}{2}}_{0}

I = \frac{1}{12}\left[sin3x\right]^{\frac{\pi}{2}}_0+\frac{3}{4}[sinx]^{\frac{\pi}{2}}_{0}

I = 1/12 [-1 – 0] + 3/4[1 – 0]

I = 3/4 – 1/12

I = (9 – 1)/12

I = 8/12

I = 2/3

Therefore, the value of \int_{0}^{\frac{\pi}{2}} cos^3xdx      is 2/3.

Question 19. \int_{0}^{\frac{\pi}{6}} cosxcos2xdx

Solution: 

We have,

I = \int_{0}^{\frac{\pi}{6}} cosxcos2xdx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{6}} 2cosxcos2xdx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{6}} (cos3x + cosx)dx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{6}}cos3xdx + \frac{1}{2}\int_{0}^{\frac{\pi}{6}}cosxdx

I = \frac{1}{2}\left[\frac{sin3x}{3}\right]^{\frac{\pi}{6}}_0 + \frac{1}{2}\left[sinx\right]^{\frac{\pi}{6}}_0

I = \frac{1}{6}\left[sin3x\right]^{\frac{\pi}{6}}_0 + \frac{1}{2}\left[sinx\right]^{\frac{\pi}{6}}_0

I = 1/6[sinπ/2 – sin0] + 1/2[sinπ/6 – sin0]

I = 1/6[1 – 0] + 1/2[1/2 – 0]

I = 1/6 + 1/4

I = (4 + 6)/24 

I = 10/24 

I = 5/12 

Therefore, the value of \int_{0}^{\frac{\pi}{6}} cosxcos2xdx      is 5/12.

Question 20. \int_{0}^{\frac{\pi}{2}} sinxsin2xdx

Solution: 

We have,

I = \int_{0}^{\frac{\pi}{2}} sinxsin2xdx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} 2sinxsin2xdx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} (cosx - cos3x)dx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}cosxdx - \frac{1}{2}\int_{0}^{\frac{\pi}{2}}cos3xdx

I = \frac{1}{2}\left[sinx\right]^{\frac{\pi}{2}}_0-\frac{1}{2}\left[\frac{sin3x}{3}\right]^{\frac{\pi}{2}}_0

I = \frac{1}{2}\left[sinx\right]^{\frac{\pi}{2}}_0-\frac{1}{6}[sin3x]^{\frac{\pi}{2}}_0

I = 1/2[sinπ/2 – sin0] – 1/6[sin3π/2 – sin0]

I = 1/2[1 – 0] – 1/6[-1 – 0]

I = 1/2 – 1/6(-1)

I = 1/2 + 1/6 

I = (6 + 2)/12

I = 8/12 

I = 2/3

Therefore, the value of \int_{0}^{\frac{\pi}{2}} sinxsin2xdx      is 2/3.

Question 21. \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (tanx+cotx)^2dx

Solution: 

We have,

I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (tanx+cotx)^2dx

I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (\frac{sinx}{cosx}+\frac{cosx}{sinx})^2dx

I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (\frac{sin^2x+cos^2x}{cosxsinx})^2dx

I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (\frac{1}{cosxsinx})^2dx

I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\frac{1}{cos^2xsin^2x}dx

I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\frac{4}{4cos^2xsin^2x}dx

I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\frac{4}{(sin2x)^2}dx

I = 4\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}cosec^22xdx

I = 4\left[\frac{-cot2x}{2}\right]_{\frac{\pi}{3}}^{\frac{\pi}{4}}

I = 2\left[-cot2x\right]_{\frac{\pi}{3}}^{\frac{\pi}{4}}

I = 2[-cotπ/2 + cot2π/3]

I = 2[-1/√3 – 0]

I = -2/√3 

Therefore, the value of \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} (tanx+cotx)^2dx      is -2/√3.

Question 22. \int_{0}^{\frac{\pi}{2}} cos^4xdx

Solution: 

We have,

I = \int_{0}^{\frac{\pi}{2}} cos^4xdx

I = \int_{0}^{\frac{\pi}{2}} (cos^2x)^2dx

I = \int_{0}^{\frac{\pi}{2}} (\frac{1+cos2x}{2})^2dx

I = \frac{1}{4}\int_{0}^{\frac{\pi}{2}} (1+cos2x)^2dx

I = \frac{1}{4}\int_{0}^{\frac{\pi}{2}} (1+cos^22x+2cos2x)dx

I = \frac{1}{4}\int_{0}^{\frac{\pi}{2}} (1+\frac{1+cos4x}{2}+2cos2x)dx

I = \frac{1}{4}\left[x+\frac{x}{2}+\frac{sin4x}{8}+sin2x\right]^{\frac{\pi}{2}}_0

I = 1/4[π/2 + π/4 + 0 + 0 – 0 – 0 – 0 – 0]

I = 1/4[3π/4] 

I = 3π/16 

Therefore, the value of \int_{0}^{\frac{\pi}{2}} cos^4xdx     is 3π/16.

Evaluate the following definite integrals:

Question 23. \int_{0}^{\frac{\pi}{2}} (a^2cos^2x+b^2sin^2x)dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} (a^2cos^2x+b^2sin^2x)dx

I = \int_{0}^{\frac{\pi}{2}} (a^2cos^2x+b^2(1-cos^2x))dx

I = \int_{0}^{\frac{\pi}{2}} [(a^2-b^2)cos^2x+b^2]dx

I = \int_{0}^{\frac{\pi}{2}} [(a^2-b^2)(\frac{1+cos2x}{2})]dx+\int_{0}^{\frac{\pi}{2}}b^2dx

I = \frac{a^2-b^2}{2}\int_{0}^{\frac{\pi}{2}} [(1+cos2x)]dx+\int_{0}^{\frac{\pi}{2}}b^2dx

I = \frac{a^2-b^2}{2}\left[x+\frac{sin2x}{2}\right]_{0}^{\frac{\pi}{2}}+b^2\left[x\right]_{0}^{\frac{\pi}{2}}

I = \frac{a^2-b^2}{2}\left[\frac{\pi}{2}+sin\pi-0-0\right]+b^2\left[\frac{\pi}{2}-0\right]

I = \frac{a^2-b^2}{2}\left[\frac{\pi}{2}\right]+b^2\left[\frac{\pi}{2}\right]

I = (\frac{a^2-b^2}{2}+b^2)\left[\frac{\pi}{2}\right]

I = (\frac{a^2-b^2+2b^2}{2})\left[\frac{\pi}{2}\right]

I = [(a2 + b2)/2][π/2]

I = π(a2 + b2)/4

Therefore, the value of \int_{0}^{\frac{\pi}{2}} (a^2cos^2x+b^2sin^2x)dx      is π(a2 + b2)/4.

Question 24. \int_{0}^{\frac{\pi}{2}} \sqrt{1+sinx}dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} \sqrt{1+sinx}dx

I = \int_{0}^{\frac{\pi}{2}} \sqrt{1+\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}dx

I = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{1+tan^2\frac{x}{2}+2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}dx

I = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{(1+tan\frac{x}{2})^2}{1+tan^2\frac{x}{2}}}dx

I = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{(1+tan\frac{x}{2})^2}{sec^2\frac{x}{2}}}dx

I = \int_{0}^{\frac{\pi}{2}} \frac{1+tan\frac{x}{2}}{sec\frac{x}{2}}dx

I = \int_{0}^{\frac{\pi}{2}} (cos\frac{x}{2}+sin\frac{x}{2})dx

I = \left[2sin\frac{x}{2}-2cos\frac{x}{2}\right]_{0}^{\frac{\pi}{2}}

I = 2[sinπ/4 – cosπ/4 – 0 + 1]

I = 2[1/√2 – 1/√2 – 0 + 1]

I = 2 (1)

I = 2

Therefore, the value of \int_{0}^{\frac{\pi}{2}} \sqrt{1+sinx}dx       is 2.

Question 25. \int_{0}^{\frac{\pi}{2}} \sqrt{1+cosx}dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} \sqrt{1+cosx}dx

I = \int_{0}^{\frac{\pi}{2}} \sqrt{2cos^2\frac{x}{2}}dx

I = \int_{0}^{\frac{\pi}{2}} \sqrt{2}cos\frac{x}{2}dx

I = \sqrt{2}\int_{0}^{\frac{\pi}{2}}cos\frac{x}{2}dx

I = \sqrt{2}\left[2sin\frac{x}{2}\right]_{0}^{\frac{\pi}{2}}

I = 2\sqrt{2}\left[sin\frac{x}{2}\right]_{0}^{\frac{\pi}{2}}

I = 2√2[sinπ/4 – sin0]

I = 2√2[1/√2- sin0]

I = 2√2[1/√2] 

I = 2

Therefore, the value of \int_{0}^{\frac{\pi}{2}} \sqrt{1+cosx}dx      is 2.

Question 26. \int_{0}^{\frac{\pi}{2}} xsinxdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} xsinxdx

By using integration by parts, we get,

I = x ∫sinxdx – ∫(∫sin x (1)dx)dx

I = -xcosx – ∫(∫sin xdx)dx

I = -xcosx + ∫cosxdx

I = -xcosx + sinx

So we get,

I = \left[-xcosx+sinx\right]^{\frac{\pi}{2}}_0

I = [-π/2cosπ/2 + sinπ/2 + 0 – 0]

I = 0 + 1 + 0 – 0 

I = 1

Therefore, the value of \int_{0}^{\frac{\pi}{2}} xsinxdx      is 1.

Question 27. \int_{0}^{\frac{\pi}{2}} xcosxdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} xcosxdx

By using integration by parts, we get,

I = x∫cosxdx – ∫(∫cos x (1)dx)dx

I = xsinx – ∫(∫cosxdx)dx

I = xsinx – ∫sinxdx

I = x sin x + cos x

So we get,

I = \left[xsinx+cosx\right]^{\frac{\pi}{2}}_0

I = [π/2sinπ/2 + cosπ/2 – 0 – cos0]

I = π/2 + 0 – 0 – 1 

I = π/2 – 1  

Therefore, the value of \int_{0}^{\frac{\pi}{2}} xcosxdx      is π/2 – 1.

Question 28. \int_{0}^{\frac{\pi}{2}} x^2cosxdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} x^2cosxdx

By using integration by parts, we get,

I = x2sinx – ∫(2x∫(cosx)dx)dx

I = x2sinx – ∫(2xsinx)dx

I = x2sinx – 2[-xcosx – ∫(1∫sinxdx)dx]

I = x2sinx – 2[-xcosx + ∫sinxdx]

I = x2sinx – 2[-xcosx + sinx]

I = x2sinx + 2xcosx – 2sinx

So we get,

I = \left[x^2sinx+2xcosx-2sinx\right]^{\frac{\pi}{2}}_0

I = [(π/2)2sinπ/2 + 2(π/2)cosπ/2 – 2sinπ/2 – 0 – 0 + sin0]

I = [π2/4 + 0 – 2 – 0 – 0 + 0]

I = π2/4 – 2 

Therefore, the value of \int_{0}^{\frac{\pi}{2}} x^2cosxdx       is π2/4 – 2.

Question 29. \int_{0}^{\frac{\pi}{4}} x^2sinxdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{4}} x^2sinxdx

By using integration by parts, we get,

I = -x2cosx – ∫(2x∫sinxdx)dx

I = -x2cosx + ∫(2xcosx)dx

I = -x2cosx + 2[xsinx – ∫(∫cosxdx)dx]

I = -x2cosx + 2[xsinx – ∫sinxdx]

I = -x2cosx + 2[xsinx + cosx]

I = -x2cosx + 2xsinx + 2cosx

So we get,

I = \left[-x^2cosx+2xsinx+2cosx\right]^{\frac{\pi}{4}}_0

I = -(π/4)2cosπ/4 + 2π/4sinπ/4 + 2cosπ/4 + 0 – 0 – 2

I = –π2/16(1/ √2) + π/2(1/√2) + 2(1/√2) + 0 – 0 – 2

I = –π2/16√2 + π/2√2 + √2 –  2

Therefore, the value of \int_{0}^{\frac{\pi}{4}} x^2sinxdx      is -π2/16√2 + π/2√2 + √2 –  2.

Question 30. \int_{0}^{\frac{\pi}{2}} x^2cos2xdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} x^2cos2xdx

By using integration by parts, we get,

I = 1/2x2sin2x – ∫(2x∫cos2xdx)dx

I = 1/2x2sin2x – ∫(xsin2x)dx

I = 1/2x2sin2x – [-1/2xcos2x – ∫(∫sin2xdx)dx]

I = 1/2x2sin2x – [-1/2xcos2x + ∫1/2 cos2xdx]

I = 1/2x2sin2x – [-1/2xcos2x + 1/4sin2xdx]

I = 1/2x2sin2x + 1/2xcos2x – 1/4sin2xdx

So we get,

I = \left[\frac{1}{2}x^2sin2x+\frac{1}{2}xcos2x-\frac{1}{4}sin2xdx\right]^{\frac{\pi}{2}}_0

I = [1/2(π2/4)sinπ + 1/2(π/2)cosπ – 0 – 0 – 0 + 0]

I = -π/4

Therefore, the value of \int_{0}^{\frac{\pi}{2}} x^2cos2xdx      is -π/4.

Question 31. \int_{0}^{\frac{\pi}{2}} x^2cos^2xdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} x^2cos^2xdx

I = \int_{0}^{\frac{\pi}{2}} x^2(\frac{1+cos2x}{2})dx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} (x^2+x^2cos2x)dx

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} x^2dx+\frac{1}{2}\int_{0}^{\frac{\pi}{2}} (x^2cos2x)dx

By using integration by parts, we get,

I = 1/2[x3/3] + x2sin2x/2 – [x ∫sin2x – ∫(∫sin2xdx)dx]

I = 1/2[x3/3] + x2sin2x/2 + xcosx/2 – sin2x/4

So we get,

I = \left[\frac{1}{2}[\frac{x^3}{3}]+\frac{x^2sin2x}{2}+\frac{xcosx}{2}-\frac{sin2x}{4}\right]^{\frac{\pi}{2}}_0

I = [1/6[π3/8] + 0 + 0 – π/8] 

I = π3/48 – π/8

Therefore, the value of \int_{0}^{\frac{\pi}{2}} x^2cos^2xdx       is π3/48 – π/8.

Question 32. \int_{1}^{2}logxdx

Solution:

We have,

I = \int_{1}^{2}logxdx

By using integration by parts, we get,

I = xlogx(1)-\int x\frac{1}{x}dx

I = xlogx – ∫1dx

I = xlogx – x

So we get,

I = \left[xlogx-x\right]^2_1

I = 2log2 – 2 – log1 + 1

I = 2 log 2 – 1

Therefore, the value of \int_{1}^{2}logxdx      is 2 log 2 – 1.

Question 33. \int_{1}^{3}\frac{logx}{(x+1)^2}dx

Solution:

We have,

I = \int_{1}^{3}\frac{logx}{(x+1)^2}dx

By using integration by parts, we get,

I = (logx)\frac{(x+1)^{-2+1}}{-2+1}-\int (\frac{1}{x}\int \frac{1}{(x+1)^2}dx)dx

I = -(x+1)^{-1}logx+\int \frac{1}{x(x+1)}dx

I = -\frac{logx}{x+1}+\int (\frac{1}{x}-\frac{1}{x+1})dx

I = -\frac{logx}{x+1}+logx - log(x+1)

So we get,

I = \left[-\frac{logx}{x+1}+logx - log(x+1)\right]^3_1

I = -log3/4 + log3 – log4 + log1/2 – log1 + log2

I = log3(1 – 1/4) – 2log2 + 0 – 0 + log2

I = 3/4log3 – log2

Therefore, the value of \int_{1}^{3}\frac{logx}{(x+1)^2}dx     is 3/4log3 – log2.

Question 34. \int_{1}^{e}\frac{e^x}{x}(1+xlogx)dx

Solution:

We have,

I = \int_{1}^{e}\frac{e^x}{x}(1+xlogx)dx

I = \int_{1}^{e}(\frac{e^x}{x}+e^xlogx)dx

I = \int_{1}^{e}\frac{e^x}{x}dx+\int_{1}^{e}e^xlogxdx

By using integration by parts, we get,

I = e^xlogx-\int_{1}^{e}e^xlogxdx+\int_{1}^{e}e^xlogxdx

I = exlogx

So we get,

I = \left[e^xlogx\right]^e_1

I = eeloge – e1log1

I = ee (1) – 0

I = ee

Therefore, the value of \int_{1}^{e}\frac{e^x}{x}(1+xlogx)dx     is ee.

Question 35. \int_{1}^{e}\frac{logx}{x}dx

Solution:

We have,

I = \int_{1}^{e}\frac{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 = e

=> t = log x

=> t = log e

=> t = 1

So, the equation becomes,

I = \int_{1}^{e}\frac{logx}{x}dx

I = \int_{0}^{1}tdt

I = \left[\frac{t^2}{2}\right]^1_0

I = 1/2 – 0/2

I = 1/2

Therefore, the value of \int_{1}^{e}\frac{logx}{x}dx     is 1/2.

Question 36. \int_{e}^{e^2}(\frac{1}{logx}-\frac{1}{(logx)^2})dx

Solution:

We have,

I = \int_{e}^{e^2}(\frac{1}{logx}-\frac{1}{(logx)^2})dx

I = \int_{e}^{e^2}\frac{1}{logx}dx-\int_{e}^{e^2}\frac{1}{(logx)^2}dx

By using integration by parts, we get,

I = \frac{x}{logx}-\int [(\frac{-1}{(logx)^2})(\frac{1}{x})\int dx]dx -\int \frac{1}{(logx)^2}dx

I = \frac{x}{logx}-\int \frac{-1}{(logx)^2}dx -\int \frac{1}{(logx)^2}dx

I = \frac{x}{logx}+\int \frac{1}{(logx)^2}dx -\int \frac{1}{(logx)^2}dx

I = x/logx

So we get,

I = \left[\frac{x}{logx}\right]^{e^2}_e

I = \left[\frac{e^2}{loge^2}-\frac{e}{loge}\right]

I = \left[\frac{e^2}{2loge}-e\right]

I = e2/2 – e

Therefore, the value of \int_{e}^{e^2}(\frac{1}{logx}-\frac{1}{(logx)^2})dx     is e2/2 – e.

Question 37. \int_{1}^{2}\frac{x+3}{x(x+2)}dx

Solution:

We have,

I = \int_{1}^{2}\frac{x+3}{x(x+2)}dx

I = \int_{1}^{2}\frac{x}{x(x+2)}dx+\int_{1}^{2}\frac{3}{x(x+2)}dx

I = \int_{1}^{2}\frac{1}{x+2}dx+\int_{1}^{2}\frac{3}{x(x+2)}dx

I = \int_{1}^{2}\frac{1}{x+2}dx+\frac{3}{2}\int_{1}^{2}(\frac{1}{x}-\frac{1}{x+2})dx

I = \left[log(x+2)\right]^2_1+\left[\frac{3}{2}logx-\frac{3}{2}log(x+2)\right]^2_1

I = \left[log(x+2)\right]^2_1+\left[\frac{3}{2}logx-\frac{3}{2}log(x+2)\right]^2_1

I = \left[\frac{3}{2}logx-\frac{1}{2}log(x+2)\right]^2_1

I = 1/2[3log2 – log4 + log3]

I = 1/2[3log2 – 2log2 + log3]

I = 1/2[log 2 – log 3]

I = 1/2[log6]

I = log6/2

Therefore, the value of \int_{1}^{2}\frac{x+3}{x(x+2)}dx     is log6/2.

Question 38. \int_{0}^{1}\frac{2x+3}{5x^2+1}dx

Solution:

We have,

I = \int_{0}^{1}\frac{2x+3}{5x^2+1}dx

I = \frac{1}{5}\int_{0}^{1}\frac{5(2x+3)}{5x^2+1}dx

I = \frac{1}{5}\int_{0}^{1}\frac{10x+15}{5x^2+1}dx

I = \frac{1}{5}\int_{0}^{1}(\frac{10x}{5x^2+1}+\frac{15}{5x^2+1}dx

I = \frac{1}{5}\int_{0}^{1}\frac{10x}{5x^2+1}+\frac{1}{5}\int_{0}^{1}\frac{15}{5x^2+1}dx

I = \frac{1}{5}\int_{0}^{1}\frac{10x}{5x^2+1}+3\int_{0}^{1}\frac{1}{5(x^2+\frac{1}{5})}dx

I = \frac{1}{5}\int_{0}^{1}\frac{10x}{5x^2+1}+\frac{3}{5}\int_{0}^{1}\frac{1}{x^2+\frac{1}{5}}dx

I = \frac{1}{5}\int_{0}^{1}\frac{10x}{5x^2+1}+\frac{3}{5}\int_{0}^{1}\frac{1}{x^2+\frac{1}{5}}dx

I = \frac{1}{5}\left[log(5x^2+1)\right]^1_0+\left[\frac{3}{5}(\frac{1}{\frac{1}{\sqrt{5}}})tan^{-1}\frac{x}{\frac{1}{\sqrt{5}}}\right]^1_0

I = \frac{1}{5}\left[log(5x^2+1)\right]^1_0+\left[\frac{3}{\sqrt{5}}tan^{-1}\sqrt{5}x\right]^1_0

I = \frac{1}{5}\left[log(5x^2+1)\right]^1_0+\left[\frac{3}{\sqrt{5}}tan^{-1}\sqrt{5}x\right]^1_0

I = [1/5log6 + 3/√5tan-1(√5) – 1/5log1 – 3/√5tan-1(0)]

I = [1/5 log6 + 3√5 tan-1(√5) – 0 – 0]

I = 1/5 log6 + 3√5 tan-1(√5)

Therefore, the value of \int_{0}^{1}\frac{2x+3}{5x^2+1}dx      is 1/5 log6 + 3√5 tan-1(√5).

Question 39. \int_{0}^{2}\frac{1}{x+4-x^2}dx

Solution:

We have,

I = \int_{0}^{2}\frac{1}{x+4-x^2}dx

I = \int_{0}^{2}\frac{1}{-(x^2+x-4)}dx

I = \int_{0}^{2}\frac{1}{-(x^2+x+\frac{1}{4}-4-\frac{1}{4})}dx

I = \int_{0}^{2}\frac{-1}{(x-\frac{1}{2})^2-\frac{17}{4}}dx

I = \int_{0}^{2}\frac{-1}{(x-\frac{1}{2})^2-(\frac{\sqrt{17}}{2})^2}dx

I = \int_{0}^{2}\frac{1}{(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2}dx

Let x – 1/2 = t, so we have,

=> dx = dt

Now, the lower limit is, x = 0

=> t = x – 1/2

=> t = 0 – 1/2

=> t = 1/2

Also, the upper limit is, x = 2

=> t = x – 1/2

=> t = 2 – 1/2

=> t = 3/2

So, the equation becomes,

I = \int_{\frac{-1}{2}}^{\frac{3}{2}}\frac{1}{(\frac{\sqrt{17}}{2})^2-t^2}dt

I = \left[\frac{1}{2(\frac{\sqrt{17}}{2})}log\frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}\right]^{\frac{3}{2}}_{\frac{-1}{2}}

I = \frac{1}{\sqrt{17}}\left[log\frac{\frac{\sqrt{17}}{2}+\frac{3}{2}}{\frac{\sqrt{17}}{2}-\frac{3}{2}}-log\frac{\frac{\sqrt{17}}{2}-\frac{1}{2}}{\frac{\sqrt{17}}{2}+\frac{1}{2}}\right]

I = \frac{1}{\sqrt{17}}\left[log\frac{\sqrt{17}+3}{\sqrt{17}-3}-log\frac{\sqrt{17}-1}{\sqrt{17}+1}\right]

I = \frac{1}{\sqrt{17}}\left[log(\frac{\sqrt{17}+3}{\sqrt{17}-3}×\frac{\sqrt{17}+1}{\sqrt{17}-1})\right]

I = \frac{1}{\sqrt{17}}\left[\log\frac{17+3+4\sqrt{17}}{17+3-4\sqrt{17}}\right]

I = \frac{1}{\sqrt{17}}\left[\log\frac{20+4\sqrt{17}}{20-4\sqrt{17}}\right]

I = \frac{1}{\sqrt{17}}\log\frac{5+\sqrt{17}}{5-\sqrt{17}}

I = \frac{1}{\sqrt{17}}\log\frac{(5+\sqrt{17})(5+\sqrt{17})}{25-17}

I = \frac{1}{\sqrt{17}}\log\frac{25+17+10\sqrt{17}}{8}

I = \frac{1}{\sqrt{17}}\log\frac{42+10\sqrt{17}}{8}

I = \frac{1}{\sqrt{17}}\log\frac{21+5\sqrt{17}}{4}

Therefore, the value of \int_{0}^{2}\frac{1}{x+4-x^2}dx      is \frac{1}{\sqrt{17}}\log\frac{21+5\sqrt{17}}{4}     .

Question 40. \int_{0}^{1}\frac{1}{2x^2+x+1}dx

Solution:

We have,

I = \int_{0}^{1}\frac{1}{2x^2+x+1}dx

I = \frac{1}{2}\int_{0}^{1}\frac{1}{x^2+\frac{x}{2}+\frac{1}{2}}dx

I = \frac{1}{2}\int_{0}^{1}\frac{1}{(x+\frac{1}{4})^2+\frac{1}{2}-\frac{1}{16}}dx

I = \frac{1}{2}\int_{0}^{1}\frac{1}{(x+\frac{1}{4})^2+\frac{7}{16}}dx

I = \frac{1}{2}\int_{0}^{1}\frac{1}{(x+\frac{1}{4})^2+(\frac{\sqrt{7}}{4})^2}dx

I = \left[\frac{1}{2}\frac{4}{\sqrt{7}}\tan^{-1}(\frac{x+\frac{1}{4}}{\frac{\sqrt{7}}{4}})\right]^1_0

I = \left[\frac{4}{2\sqrt{7}}\tan^{-1}(\frac{x+\frac{1}{4}}{\frac{\sqrt{7}}{4}})\right]^1_0

I = \frac{4}{2\sqrt{7}}\left[\tan^{-1}(\frac{\frac{5}{4}}{\frac{\sqrt{7}}{4}})-\tan^{-1}(\frac{\frac{1}{4}}{\frac{\sqrt{7}}{4}})\right]

I = 4/2√7[tan-1(5/√7) – tan-1(1/√7)]

I = 2/√7[tan-1(5/√7) – tan-1(1/√7)]

Therefore, the value of \int_{0}^{1}\frac{1}{2x^2+x+1}dx     is 2/√7[tan-1(5/√7) – tan-1(1/√7)].

Question 41. \int_{0}^{1}\sqrt{x(1-x)}dx

Solution:

We have,

I = \int_{0}^{1}\sqrt{x(1-x)}dx

Let x = sin2 t, so we have,

=> dx = 2 sin t cos t dt

Now, the lower limit is, x = 0

=> sin2 t = 0

=> sin t = 0

=> t = 0

Also, the upper limit is, x = 1

=> sin2 t = 1

=> sin t = 1

=> t = π/2

So, the equation becomes,

I = \int_{0}^{\frac{\pi}{2}}\sqrt{sin^2t(1-sin^2t)}(2sintcost)dt

I = \int_{0}^{\frac{\pi}{2}}\sqrt{sin^2t(cos^2t)}(2sintcost)dt

I = \int_{0}^{\frac{\pi}{2}}(sintcost)(2sintcost)dt

I = \int_{0}^{\frac{\pi}{2}}(2sin^2tcos^2t)dt

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}(4sin^2tcos^2t)dt

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}(sin^22t)dt

I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}(\frac{1-cos4t}{2})dt

I = \frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1-cos4t)dt

I = \frac{1}{4}\int_{0}^{\frac{\pi}{2}}dt-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}(cos4t)dt

I = \frac{1}{4}\left[t\right]^{\frac{\pi}{2}}_0-\frac{1}{4}\left[\frac{sin4t}{4}\right]^{\frac{\pi}{2}}_0

I = \frac{1}{4}\left[t\right]^{\frac{\pi}{2}}_0-\frac{1}{16}\left[sin4t\right]^{\frac{\pi}{2}}_0

I = 1/4[π/2 – 0] – 1/16[sin2π – 0]

I = 1/4[π/2] – 1/16[0 – 0 ]

I = π/8

Therefore, the value of \int_{0}^{1}\sqrt{x(1-x)}dx     is π/8.

Question 42. \int_{0}^{2}\frac{1}{\sqrt{3+2x-x^2}}dx

Solution:

We have,

I = \int_{0}^{2}\frac{1}{\sqrt{3+2x-x^2}}dx

I = \int_{0}^{2}\frac{1}{\sqrt{3+1-(x^2-2x+1)}}dx

I = \int_{0}^{2}\frac{1}{\sqrt{4-(x^2-2x+1)}}dx

I = \int_{0}^{2}\frac{1}{\sqrt{(2)^2-(x-1)^2}}dx

I = \left[sin^{-1}(\frac{x-1}{2})\right]_{0}^{2}

I = [sin-1(1/2) – sin-1(-1/2)]

I = π/6 -(-π/6)

I = π/6 + π/6

I = π/3

Therefore, the value of \int_{0}^{2}\frac{1}{\sqrt{3+2x-x^2}}dx      is π/3.

Question 43. \int_{0}^{4}\frac{1}{\sqrt{4x-x^2}}dx

Solution:

We have,

I = \int_{0}^{4}\frac{1}{\sqrt{4x-x^2}}dx

I = \int_{0}^{4}\frac{1}{\sqrt{4-4+4x-x^2}}dx

I = \int_{0}^{4}\frac{1}{\sqrt{4-(x^2-4x+4)}}dx

I = \int_{0}^{4}\frac{1}{\sqrt{2^2-(x-2)^2}}dx

I = \left[sin^{-1}(\frac{x-2}{2})\right]^4_0

I = \left[sin^{-1}(\frac{4-2}{2})-sin^{-1}(\frac{0-2}{2})\right]

I = [sin-1(2/2) – sin-1(-2/2)]

I = sin-11 – sin-1(-1)

I = π/2 – (-π/2) 

I = π/2 + π/2 

I = π

Therefore, the value of \int_{0}^{4}\frac{1}{\sqrt{4x-x^2}}dx      is π.

Question 44. \int_{-1}^{1}\frac{1}{x^2+2x+5}dx

Solution:

We have,

I = \int_{-1}^{1}\frac{1}{x^2+2x+5}dx

I = \int_{-1}^{1}\frac{1}{x^2+2x+1+4}dx

I = \int_{-1}^{1}\frac{1}{(x+1)^2+2^2}dx

Let x + 1 = t, so we have,

=> dx = dt

Now, the lower limit is, x = –1

=> t = x + 1

=> t = – 1 + 1

=> t = 0

Also, the upper limit is, x = 1

=> t = x + 1

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I = \int_{0}^{2}\frac{1}{t^2+2^2}dt

I = \left[\frac{1}{2}tan^{-1}\frac{t}{2}\right]^2_0

I = 1/2tan-12/2 – 1/2tan-10/2

I = 1/2tan-11 – 1/2tan-10

I = 1/2(π/4) – 0

I = π/8

Therefore, the value of \int_{-1}^{1}\frac{1}{x^2+2x+5}dx     is π/8.

Evaluate the following definite integrals:

Question 45. \int_{1}^{4} \frac{x^2+x}{\sqrt{2x+1}}dx

Solution:

We have,

I = \int_{1}^{4} \frac{x^2+x}{\sqrt{2x+1}}dx

Let 2x + 1 = t2, so we have,

=> 2 dx = 2t dt

=> dx = t dt

Now, the lower limit is, x = 1

=> t2 = 2x + 1

=> t2 = 2(1) + 1

=> t2 = 3

=> t = √3 

Also, the upper limit is, x = 4

=> t= 2x + 1

=> t2 = 2(4) + 1

=> t2 = 9

=> t = 3 

So, the equation becomes,

I = \int_{\sqrt{3}}^{3} \frac{(\frac{t^2-1}{2})^2+(\frac{t^2-1}{2})}{t}tdt

I = \int_{\sqrt{3}}^{3} (\frac{t^2-1}{2})^2+(\frac{t^2-1}{2})dt

I = \frac{1}{4}\int_{\sqrt{3}}^{3} (t^4+1-2t^2+2t^2-2)dt

I = \frac{1}{4}\int_{\sqrt{3}}^{3} (t^4-1)dt

I = \frac{1}{4}\left[\frac{t^5}{5}-t\right]^3_{\sqrt{3}}

I = 1/4 [35/5 – 3 – (√3)5/5 + √3]

I = 1/4[243/5 – 3 – 9√3/5 + √3]

I = 1/4((243 – 15 – 9√3 + 5√3)/5)

I = 1/4[(228 – 4√3)/5]

I = 1/4[4(57 – √3)/5]

I = (57 – √3)/5 

Therefore, the value of \int_{1}^{4} \frac{x^2+x}{\sqrt{2x+1}}dx    is (57 – √3)/5.

Question 46. \int_{0}^{1} x(1-x)^5dx

Solution:

We have,

I = \int_{0}^{1} x(1-x)^5dx

By using binomial theorem in the expansion of (1 – x)5, we get,

I = \int_{0}^{1} x[1^5+^5C_1(-x)+^5C_2(-x)^2+^5C_3(-x)^3+^5C_4(-x)^4+^5C_5(-x)^5]dx

I = \int_{0}^{1} x(1-5x+10x^2-10x^3+5x^4-x^5)dx

I = \int_{0}^{1} (x-5x^2+10x^3-10x^4+5x^5-x^6)dx

I = \left[\frac{x^2}{2}-\frac{5x^3}{3}+\frac{10x^4}{4}-\frac{10x^5}{5}+\frac{5x^6}{6}-\frac{x^7}{7}\right]^{1}_0

I = 1/2 – 5/3 + 10/4 – 10/5 + 5/6 – 1/7

I = 1/2 – 5/3 + 5/3 – 2 + 5/6 – 1/7

I = 1/2 – 2 + 5/6 – 1/7 

I = 1/42

Therefore, the value of \int_{0}^{1} x(1-x)^5dx     is 1/42.

Question 47. \int_{1}^{2} (\frac{x-1}{x^2})e^xdx

Solution:

We have,

I = \int_{1}^{2} (\frac{x-1}{x^2})e^xdx

I = \int_{1}^{2} \frac{xe^x-e^x}{x^2}dx

I = \int_{1}^{2} (\frac{e^x}{x}-\frac{e^x}{x^2})dx

I = \int_{1}^{2}\frac{e^x}{x}dx-\int_{1}^{2}\frac{e^x}{x^2}dx

By using integration by parts, we get,

I = \frac{e^x}{x}-\int ((\frac{-1}{x^2})\int e^xdx)dx-\int \frac{e^x}{x^2}dx

I = \frac{e^x}{x}+\int\frac{e^x}{x^2}dx-\int \frac{e^x}{x^2}dx

I = ex/x

So we get,

I = \left[\frac{e^x}{x}\right]^2_1

I = e2/2 – e1/1

I = e2/2 – e

Therefore, the value of \int_{1}^{2} (\frac{x-1}{x^2})e^xdx    is e2/2 – e.

Question 48. \int_{0}^{1} (xe^{2x}+sin\frac{\pi x}{2})dx

Solution:

We have,

I = \int_{0}^{1} (xe^{2x}+sin\frac{\pi x}{2})dx

By using integration by parts in first integral, we get,

I = \frac{xe^{2x}}{2}-\frac{1}{2}\int e^{2x}dx+\left[\frac{-cos\frac{\pi x}{2}}{\frac{\pi}{2}}\right]^1_0

I = xe2x/2 – (1/2)(e2x/2) + 2/π[1 – 0]

I = xe2x/2 – e2x/4 + 2/π

So we get,

I = \left[\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\right]^1_0+\frac{2}{\pi}

I = [e2/2 + e2/4 – 0 + 1/4] + 2/π

I = e2/4 + 1/4 + 2/π

Therefore, the value of \int_{0}^{1} (xe^{2x}+sin\frac{\pi x}{2})dx     is e2/4 + 1/4 + 2/π.

Question 49. \int_{0}^{1} (xe^{x}+cos\frac{\pi x}{4})dx

Solution:

We have,

I = \int_{0}^{1} (xe^{x}+cos\frac{\pi x}{4})dx

By using integration by parts in first integral, we get,

I = xe^{x}-\int ((1)\int e^xdx)dx+ \left[\frac{sin\frac{\pi x}{4}}{\frac{\pi x}{4}}\right]^1_0

I = xe^{x}-\int e^xdx+\frac{4}{\pi}[\frac{1}{\sqrt{2}}-0]

I = xe^{x}-e^x+\frac{4}{\pi}[\frac{1}{\sqrt{2}}]

I = xe^{x}-e^x+\frac{2\sqrt{2}}{\pi}

So we get,

I = \left[xe^{x}-e^x\right]^1_0+\frac{2\sqrt{2}}{\pi}

I = \left[e^{x}(x-1)\right]^1_0+\frac{2\sqrt{2}}{\pi}

I = [e1(1 – 1) – e0(0 – 1)] + 2√2/π

I = [0 – (-1)] + 2√2/π

I = 1 + 2√2/π

Therefore, the value of \int_{0}^{1} (xe^{x}+cos\frac{\pi x}{4})dx     is 1 + 2√2/π.

Question 50. \int_{\frac{\pi}{2}}^{\pi} e^{x}(\frac{1-sinx}{1-cosx})dx

Solution:

We have,

I = \int_{\frac{\pi}{2}}^{\pi} e^{x}(\frac{1-sinx}{1-cosx})dx

I = \int_{\frac{\pi}{2}}^{\pi} e^{x}(\frac{1-2sin\frac{x}{2}cos\frac{x}{2}}{2sin^2\frac{x}{2}})dx

I = -\int_{\frac{\pi}{2}}^{\pi} e^{x}(\frac{-1}{2}cosec^2\frac{x}{2}+cot\frac{x}{2})dx

I = \left[-e^xcot\frac{x}{2}\right]^\pi_\frac{\pi}{2}

I = -eπ cotπ/2 + eπ/2 cotπ/4

I = 0 + eπ/2(1)

I = eπ/2

Therefore, the value of \int_{\frac{\pi}{2}}^{\pi} e^{x}(\frac{1-sinx}{1-cosx})dx    is eπ/2.

Question 51. \int_{0}^{2\pi} e^{\frac{x}{2}}\sin(\frac{x}{2}+\frac{\pi}{4})dx

Solution:

We have,

I = \int_{0}^{2\pi} e^{\frac{x}{2}}\sin(\frac{x}{2}+\frac{\pi}{4})dx

I = \int_{0}^{2\pi} e^{\frac{x}{2}}(sin\frac{x}{2}cos\frac{\pi}{4}+cos\frac{x}{2}sin\frac{\pi}{4})dx

I = \int_{0}^{2\pi} e^{\frac{x}{2}}(\frac{1}{\sqrt{2}}sin\frac{x}{2}+\frac{1}{\sqrt{2}}cos\frac{x}{2})dx

I = \int_{0}^{2\pi} e^{\frac{x}{2}}(\frac{1}{\sqrt{2}}sin\frac{x}{2})dx+\int_{0}^{2\pi} e^{\frac{x}{2}}(\frac{1}{\sqrt{2}}cos\frac{x}{2})dx

By using integration by parts in first integral, we get,

I = \frac{1}{\sqrt{2}}\left[sin\frac{x}{2}\int_{0}^{2\pi} e^{\frac{x}{2}}dx-\int (\int (e^{\frac{x}{2}}dx)\frac{sin\frac{x}{2}}{2})dx\right]+\int_{0}^{2\pi} e^{\frac{x}{2}}(\frac{1}{\sqrt{2}}cos\frac{x}{2})dx

I = \frac{1}{\sqrt{2}}\left[sin\frac{x}{2}(2e^{\frac{x}{2}})\right]^{2\pi}_0-\int_{0}^{2\pi}e^{\frac{x}{2}}(\frac{1}{\sqrt{2}}cos\frac{x}{2})dx+\int_{0}^{2\pi}e^{\frac{x}{2}}(\frac{1}{\sqrt{2}}cos\frac{x}{2})dx

I = \frac{1}{\sqrt{2}}\left[sin\frac{x}{2}(2e^{\frac{x}{2}})\right]^{2\pi}_0

I = 1/√2[sinπ(2eπ) – 0]

I = 1/√2[0 – 0]

I = 0

Therefore, the value of \int_{0}^{2\pi} e^{\frac{x}{2}}\sin(\frac{x}{2}+\frac{\pi}{4})dx    is 0.

Question 52. \int_{0}^{2\pi} e^x\cos(\frac{x}{2}+\frac{\pi}{4})dx

Solution:

We have,

I = \int_{0}^{2\pi} e^x\cos(\frac{x}{2}+\frac{\pi}{4})dx

By using integration by parts, we get,

I = excos(x/2 + π/4) + 1/2∫exsin(x/2 + π/4)

I = ecos(x/2 + π/4) + 1/2[ exsin(x/2 + π/4) – 1/2 ∫excos(x/2 + π/4)dx]

I = excos(x/2 + π/4) + 1/2exsin(x/2 + π/4) – 1/4I

\frac{5I}{4}=\left[e^x\cos(\frac{x}{2}+\frac{\pi}{4})+\frac{1}{2}e^xsin(\frac{x}{2}+\frac{\pi}{4})\right]^{2\pi}_0
\frac{5I}{4}=\left[\frac{-1}{\sqrt{2}}(e^{2\pi}+1)-\frac{1}{2\sqrt{2}}(e^{2\pi}+1)\right]

5I/4 = -3/ 2√2(e2π + 1)

I = -3√2/5(e + 1)

Therefore, the value of \int_{0}^{2\pi} e^x\cos(\frac{x}{2}+\frac{\pi}{4})dx    is -3√2/5(e + 1).

Question 53. \int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}}

Solution:

We have,

I = \int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}}

I = \int_{0}^{1} \frac{\sqrt{1+x}+\sqrt{x}}{(\sqrt{1+x}-\sqrt{x})(\sqrt{1+x}+\sqrt{x})}dx

I = \int_{0}^{1} \frac{\sqrt{1+x}+\sqrt{x}}{1+x-x}dx

I = \int_{0}^{1} (\sqrt{1+x}+\sqrt{x})dx

I = \int_{0}^{1}(\sqrt{1+x})dx+\int_{0}^{1}\sqrt{x}dx

I = \left[\frac{2}{3}(1+x)^{\frac{3}{2}}\right]^1_0+\left[\frac{2}{3}x^{\frac{3}{2}}\right]^1_0

I = 2/3[23/2 – 1] + 2/3[1 – 0]

I = (\frac{2^{1+\frac{3}{2}}}{3})-\frac{2}{3}+\frac{2}{3}

I = 25/2/3 

Therefore, the value of \int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}}    is 25/2/3.

Question 54. \int_{1}^{2} \frac{x}{(x+1)(x+2)}dx

Solution:

We have,

I = \int_{1}^{2} \frac{x}{(x+1)(x+2)}dx

I = -\int_{1}^{2} \frac{1}{x+1}dx+\int_{1}^{2} \frac{2}{x+2}dx

I = -\left[log(x+1)\right]^2_1+\left[2log(x+2)\right]^2_1

I = -\left[log(x+1)\right]^2_1+2\left[log(x+2)\right]^2_1

I = -log3 + log2 + 2[log4 – log3]

I = -log3 + log2 + 2[2log2 – log3]

I = -log3 + log2 + 4log2 – 2log3

I = 5log2 – 3log3

I = log2– log33

I = log32 – log27

I = log32/27 

Therefore, the value of \int_{1}^{2} \frac{x}{(x+1)(x+2)}dx    is log32/27.

Question 55. \int_{0}^{\frac{\pi}{2}} sin^3xdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}} sin^3xdx

I = \int_{0}^{\frac{\pi}{2}} sin^2x(sinx)dx

I = \int_{0}^{\frac{\pi}{2}} (1-cos^2x)(sinx)dx

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} (t^2-1)dt

I = \int_{1}^{0}t^2dt-\int_{1}^{0}(1)dt

I = \left[\frac{t^2}{3}\right]^{0}_1-\left[t\right]^0_1

I = [0 – 1/3] – [0 – 1]

I = [-1/3] – [-1]

I = -1/3 + 1

I = 2/3

Therefore, the value of \int_{0}^{\frac{\pi}{2}} sin^3xdx    is 2/3.

Question 56. \int_{0}^{\pi} (sin^2\frac{x}{2}-cos^2\frac{x}{2})dx

Solution:

We have,

I = \int_{0}^{\pi} (sin^2\frac{x}{2}-cos^2\frac{x}{2})dx

I = -\int_{0}^{\pi} (cos^2\frac{x}{2}-sin^2\frac{x}{2})dx

I = -\int_{0}^{\pi}cosxdx

I = \left[-sinx\right]^\pi_0

I = -sinπ + sin0 

I = 0

Therefore, the value of \int_{0}^{\pi} (sin^2\frac{x}{2}-cos^2\frac{x}{2})dx    is 0.

Question 57. \int_{1}^{2}(\frac{1}{x}-\frac{1}{2x^2})e^{2x}dx

Solution:

 We have,

I = \int_{1}^{2}(\frac{1}{x}-\frac{1}{2x^2})e^{2x}dx

Let 2x = t, so we have,

=> 2x dx = dt

Now, the lower limit is, x = 1

=> t = 2x

=> t = 2(1)

=> t = 2

Also, the upper limit is, x =  2

=> t = 2x

=> t = 2(2)

=> t = 4

So, the equation becomes,

I = \frac{1}{2}\int_{2}^{4}(\frac{2}{t}-\frac{4}{2t^2})e^{t}dt

I = \int_{2}^{4}(\frac{1}{t}-\frac{1}{t^2})e^{t}dt

I = \int_{2}^{4}\frac{e^t}{t}dt-\int_{2}^{4}\frac{e^t}{t^2}dt

By using integration by parts in first integral, we get,

I = \left[\frac{e^t}{t}\right]^4_2-\int (\frac{-1}{t^2}\int e^tdx)dx-\int_{2}^{4}\frac{e^t}{t^2}dt

I = \left[\frac{e^t}{t}\right]^4_2+\int_{2}^{4}\frac{e^t}{t^2}dt-\int_{2}^{4}\frac{e^t}{t^2}dt

I = \left[\frac{e^t}{t}\right]^4_2

I = e4/4 – e2/2

Therefore, the value of \int_{1}^{2}(\frac{1}{x}-\frac{1}{2x^2})e^{2x}dx    is e4/4 – e2/2.

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

Solution:

We have,

I = \int_{1}^{2}\frac{1}{\sqrt{(x-1)(2-x)}}dx

I = \int_{1}^{2}\frac{1}{\sqrt{2x-x^2-2+x}}dx

I = \int_{1}^{2}\frac{1}{\sqrt{-x^2+3x-2}}dx

I = \int_{1}^{2}\frac{1}{\sqrt{-(x-\frac{3}{2})^2+\frac{1}{4}}}dx

I = \int_{1}^{2}\frac{1}{\sqrt{(\frac{1}{2})^2-(x-\frac{3}{2})^2}}dx

I = \left[sin^{-1}(2x-3)\right]_{1}^{2}

I = [sin-1(1) – sin-1(-1)]

I = π/2 – (-π/2)

I = π/2 + π/2

I =  π

Therefore, the value of \int_{1}^{2}\frac{1}{\sqrt{(x-1)(2-x)}}dx    is π.

Question 59. If \int_{0}^{k}\frac{1}{2+8x^2}dx=\frac{\pi}{16}   , find the value of k.

Solution:

We have,

=> \int_{0}^{k}\frac{1}{2+8x^2}dx=\frac{\pi}{16}

=> \int_{0}^{k}\frac{1}{2(1+4x^2)}dx=\frac{\pi}{16}

=> \int_{0}^{k}\frac{1}{2(1+(2x)^2)}dx=\frac{\pi}{16}

=> \left[\frac{tan^{-1}2x}{4}\right]_{0}^{k}=\frac{\pi}{16}

=> tan-12k/4 – tan-10 = π/16

=> tan-12k/4 – 0 = π/16

=> tan-12k/4 = π/16

=> tan-12k = π/4

=> 2k = tanπ/4

=> 2k = 1

=> k = 1/2

Therefore, the value of k is 1/2.

Question 60. If \int_{0}^{a}3x^2dx=8   , find the value of k.

Solution:

We have,

=> \int_{0}^{a}3x^2dx=8

=> \left[\frac{3(x^{2+1})}{2+1}\right]^a_0=8

=> \left[\frac{3(x^{3})}{3}\right]^a_0=8

=> \left[x^3\right]^a_0=8

=> a3 – 0 = 8

=> a3 = 8

=> a = 2

Therefore, the value of a is 2.

Question 61. \int_\pi^\frac{3\pi}{2}\sqrt{1-cos2x}dx    

Solution:

We have,

I = \int_\pi^\frac{3\pi}{2}\sqrt{1-cos2x}dx

I = \int_\pi^\frac{3\pi}{2}\sqrt{2sin^2x}dx

I = \int_\pi^\frac{3\pi}{2}\sqrt{2}sinxdx

I = -\left[\sqrt{2}cosx\right]_\pi^\frac{3\pi}{2}

I = -[√2cos3π/2 – √2cosπ]

I = -(-√2 – 0)

I = √2

Therefore, the value of \int_\pi^\frac{3\pi}{2}\sqrt{1-cos2x}dx    is √2.

Question 62. \int_0^{2\pi}\sqrt{1+sin\frac{x}{2}}dx

Solution:

We have,

I = \int_0^{2\pi}\sqrt{1+sin\frac{x}{2}}dx

I = \int_0^{2\pi}\sqrt{sin^2\frac{x}{4}+cos^2\frac{x}{4}+2sin\frac{x}{4}cos\frac{x}{4}}dx

I = \int_0^{2\pi}\sqrt{(sin\frac{x}{4}+cos\frac{x}{4})^2}dx

I = \int_0^{2\pi}(sin\frac{x}{4}+cos\frac{x}{4})dx

I = \int_0^{2\pi}sin\frac{x}{4}dx+\int_0^{2\pi}cos\frac{x}{4}dx

I = \left[\frac{-cos\frac{x}{4}}{\frac{1}{4}}\right]_0^{2\pi}+\left[\frac{sin\frac{x}{4}}{\frac{1}{4}}\right]_0^{2\pi}

I = \left[-4cos\frac{x}{4}\right]_0^{2\pi}+\left[4sin\frac{x}{4}\right]_0^{2\pi}

I = [-4cosπ/2 + 4cos0] + [4sinπ/2 – 4sin0]

I = 0 + 4 + 4 – 0

I = 8

Therefore, the value of \int_0^{2\pi}\sqrt{1+sin\frac{x}{2}}dx    is 8.

Question 63. \int_0^{\frac{\pi}{4}}(tanx+cotx)^{-2}dx

Solution:

We have,

I = \int_0^{\frac{\pi}{4}}(tanx+cotx)^{-2}dx

I = \int_0^{\frac{\pi}{4}}\frac{1}{(tanx+cotx)^2}dx

I = \int_0^{\frac{\pi}{4}}\frac{1}{(\frac{sinx}{cosx}+\frac{cosx}{sinx})^2}dx

I = \int_0^{\frac{\pi}{4}}\frac{1}{(\frac{sin^2x+cos^2x}{sinxcosx})^2}dx

I = \int_0^{\frac{\pi}{4}}\frac{1}{(\frac{1}{sinxcosx})^2}dx

I = \int_0^{\frac{\pi}{4}}sin^2xcos^2xdx

I = \int_0^{\frac{\pi}{4}}sin^2x(1-sin^2x)dx

I = \int_0^{\frac{\pi}{4}}(sin^2x-sin^4x)dx

I = \int_0^{\frac{\pi}{4}}sin^2xdx-\int_0^{\frac{\pi}{4}}sin^4xdx

I = \left[\frac{x}{2}-\frac{cosxsinx}{2}\right]^{\frac{\pi}{4}}_0-\left[\frac{3}{4}(\frac{x}{2}-\frac{cosxsinx}{2})-\frac{cosxsin^3x}{4}\right]^{\frac{\pi}{4}}_0

I = (\frac{\pi}{8}-\frac{cos\frac{\pi}{4}sin\frac{\pi}{4}}{2})-\left(\frac{3}{4}(\frac{\pi}{8}-\frac{cos\frac{\pi}{4}sin\frac{\pi}{4}}{2})-\frac{cos\frac{\pi}{4}sin^3\frac{\pi}{4}}{4}\right)

I = (π/8 – 1/4) – (3/4(π/8 – 1/4) – 1/16)

I = π/8 – 1/4 – (3π/32 – 3/16 – 1/16)

I =  π/8 – 1/4 – (3π/32 – 1/4)

I = π/8 – 1/4 – 3π/32 + 1/4

I = π/8 – 3π/32

I = (4π – 3π)/32

I = π/32 

Therefore, the value of \int_0^{\frac{\pi}{4}}(tanx+cotx)^{-2}dx    is π/32.

Question 64. \int_0^{1}xlog(1+2x)dx

Solution:

We have,

I = \int_0^{1}xlog(1+2x)dx

By using integration by parts we get,

I = \frac{x^2log(1+2x)}{2}-\int(\frac{2}{2x+1}\int xdx)dx

I = \frac{x^2log(1+2x)}{2}-\int \frac{2x^2}{2(2x+1)}dx

I = \frac{x^2log(1+2x)}{2}-\int \frac{x^2}{2x+1}dx

I = \frac{x^2log(1+2x)}{2}-\int_0^1 \frac{x}{2}-\frac{1}{4}+\frac{1}{4(2x+1)}dx

I = \frac{x^2log(1+2x)}{2}-(\frac{x^2}{4}-\frac{x}{4}+\frac{1}{8}log|2x+1|)

So we get,

I = \left[\frac{x^2log(1+2x)}{2}\right]^1_0-\left[\frac{x^2}{4}-\frac{x}{4}+\frac{1}{8}log|2x+1|)\right]^1_0

I = log3/2 – 1/8log3

I = 3/8log3

Therefore, the value of \int_0^{1}xlog(1+2x)dx    is 3/8log3.

Question 65. \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(tanx+cotx)^{2}dx

Solution:

We have,

I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(tanx+cotx)^{2}dx

I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(tan^2x+cot^2x+2tanxcotx)dx

I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(sec^2x-1+cosec^2x-1+2)dx

I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(sec^2x+cosec^2x-2+2)dx

I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(sec^2x+cosec^2x)dx

I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}sec^2xdx+\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}cosec^2xdx

I = \left[tanx\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}+\left[-cotx\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}

I = [tanπ/3 – tanπ/6] + [-cotπ/3 + cotπ/6]

I = [√3 – 1/√3] + [- 1/√3 – √3]

I = 2[√3 – 1/√3]

I = 4/√3

Therefore, the value of \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(tanx+cotx)^{2}dx    is 4/√3.

Question 66. \int_{0}^{\frac{\pi}{4}}(a^2cos^2x+b^2sin^2x)dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{4}}(a^2cos^2x+b^2sin^2x)dx

I = \int_{0}^{\frac{\pi}{4}}(a^2(1-sin^2x)+b^2sin^2x)dx

I = \int_{0}^{\frac{\pi}{4}}(a^2-a^2sin^2x+b^2sin^2x)dx

I = \int_{0}^{\frac{\pi}{4}}[a^2-(b^2-a^2)sin^2x]dx

I = \int_{0}^{\frac{\pi}{4}}a^2dx-\int_{0}^{\frac{\pi}{4}}[(b^2-a^2)sin^2x]dx

I = \int_{0}^{\frac{\pi}{4}}a^2dx-(b^2-a^2)\int_{0}^{\frac{\pi}{4}}(\frac{1+cos2x}{2})dx

I = \int_{0}^{\frac{\pi}{4}}a^2dx-(b^2-a^2)\int_{0}^{\frac{\pi}{4}}(\frac{1+cos2x}{2})dx

I = \left[a^2x\right]_{0}^{\frac{\pi}{4}}+(b^2-a^2)[\frac{x}{2}+\frac{sin2x}{4}]_{0}^{\frac{\pi}{4}}

I = \frac{a^2\pi}{4}+(b^2-a^2)(\frac{\pi}{8}+\frac{1}{4})

I = \frac{a^2\pi}{4}-\frac{(b^2-a^2)\pi}{8}+\frac{b^2-a^2}{4}

I = \frac{(b^2-a^2)\pi}{8}+\frac{b^2-a^2}{4}

Therefore, the value of \int_{0}^{\frac{\pi}{4}}(a^2cos^2x+b^2sin^2x)dx    is \frac{(b^2-a^2)\pi}{8}+\frac{b^2-a^2}{4}   .

Question 67. \int_{0}^{1}\frac{dx}{1+2x+2x^2+2x^3+x^4}

Solution:

We have,

I = \int_{0}^{1}\frac{dx}{1+2x+2x^2+2x^3+x^4}

I = \int_{0}^{1}\frac{dx}{(x+1)^2(x^2+1)}

I = \int_{0}^{1}[\frac{-x}{2(x^2+1)}+\frac{1}{2(x+1)}+\frac{1}{2(x+1)^2}]dx

I = -\left[\frac{log(x^2+1)}{4}\right]^1_0+\left[\frac{log(x+1)}{2}\right]^1_0-\left[\frac{1}{2(x+1)}\right]^1_0

I = -log2/4 + log2/2 – 1/4 + 1/2

I = log2/4 + 1/4

Therefore, the value of \int_{0}^{1}\frac{dx}{1+2x+2x^2+2x^3+x^4}    is log2/4 + 1/4.

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