Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.2 | Set 3

Last Updated : 30 Jun, 2021

Evaluate the following definite integrals:

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

Solution:

We have,

I = $\int_{0}^{\pi}5(5-4\cos\theta)^{\frac{1}{4}}\sin\theta d\theta$

Let 5 – 4 cos θ = t. So, we have

=> 4 sin θ dθ = dt

=> sin θ dθ = dt/4

Now, the lower limit is, θ = 0

=> t = 5 – 4 cos θ

=> t = 5 – 4 cos 0

=> t = 5 – 4

=> t = 1

Also, the upper limit is, θ = π

=> t = 5 – 4 cos θ

=> t = 5 – 4 cos π

=> t = 5 + 4

=> t = 9

So, the equation becomes,

I = $\int_{1}^{9}\frac{5t^{\frac{1}{4}}}{4}dt$

I = $\frac{5}{4}\int_{1}^{9}t^{\frac{1}{4}}dt$

I = $\frac{5}{4}\left[\frac{t^{\frac{5}{4}}}{\frac{5}{4}}\right]_{1}^{9}$

I = $\frac{5}{4}\left[\frac{4}{5}t^{\frac{5}{4}}\right]_{1}^{9}$

I = $\left[t^{\frac{5}{4}}\right]_{1}^{9}$

I = $9^{\frac{5}{4}}-1^{\frac{5}{4}}$

I = $3^{\frac{5}{2}}-1$

I = 9√3 – 1

Therefore, the value of $\int_{0}^{\pi}5(5-4\cos\theta)^{\frac{1}{4}}\sin\theta d\theta$ is 9√3 – 1.

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

Solution:

We have,

I = $\int_{0}^{\frac{\pi}{6}}\cos^{-3}2\theta \sin2\theta d\theta$

I = $\int_{0}^{\frac{\pi}{6}}\frac{\sin2\theta}{\cos^{3}2\theta} d\theta$

I = $\int_{0}^{\frac{\pi}{6}}\frac{\sin2\theta}{(\cos2\theta)(\cos^22\theta)} d\theta$

I = $\int_{0}^{\frac{\pi}{6}}\frac{\tan2\theta}{\cos^22\theta} d\theta$

I = $\int_{0}^{\frac{\pi}{6}}\tan2\theta \sec^22\theta d\theta$

Let tan 2θ = t. So, we have

=> 2 sec² 2θ dθ = dt

=> sec² 2θ dθ = dt/2

Now, the lower limit is, θ = 0

=> t = tan 2θ

=> t = tan 0

=> t = 0

Also, the upper limit is, θ = π/6

=> t = tan 2θ

=> t = tan π/3

=> t = √3

So, the equation becomes,

I = $\frac{1}{2}\int_{0}^{\sqrt{3}}tdt$

I = $\frac{1}{2}\left[\frac{t^2}{2}\right]_{0}^{\sqrt{3}}$

I = $\frac{1}{2}\left[\frac{3}{2}-0\right]$

I = $\frac{3}{4}$

Therefore, the value of $\int_{0}^{\frac{\pi}{6}}\cos^{-3}2\theta \sin2\theta d\theta$ is $\frac{3}{4}$ .

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

Solution:

We have,

I = $\int_{0}^{\pi^\frac{3}{2}}\sqrt{x}cos^2x^\frac{2}{3}dx$

Let $x^{\frac{2}{3}}$ = t. So, we have

=> $\frac{3}{2}\sqrt{x}dx$ = dt

Now, the lower limit is, x = 0

=> t = $x^{\frac{2}{3}}$

=> t = $0^{\frac{2}{3}}$

=> t = 0

Also, the upper limit is, x = $\pi^{\frac{3}{2}}$

=> t = $x^{\frac{2}{3}}$

=> t = $(\pi^{\frac{3}{2}})^{\frac{2}{3}}$

=> t = π

So, the equation becomes,

I = $\frac{2}{3}\int_{0}^{\pi}cos^2tdt$

I = $\frac{1}{3}\int_{0}^{\pi}(1+cos2t)dt$

I = $\frac{1}{3}\left[t+\frac{sin2t}{t}\right]_{0}^{\pi}$

I = $\frac{1}{3}\left[\pi+0-0-0\right]$

I = $\frac{\pi}{3}$

Therefore, the value of $\int_{0}^{\pi^\frac{3}{2}}\sqrt{x}cos^2x^\frac{2}{3}dx$ is $\frac{\pi}{3}$ .

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

Solution:

We have,

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

Let 1 + log x = t. So, we have

=> 1/x dx = dt

Now, the lower limit is, x = 0

=> t = 1 + log x

=> t = 1 + log 0

=> t = 1

Also, the upper limit is, x = 2

=> t = 1 + log x

=> t = 1 + log 2

So, the equation becomes,

I = $\int_{1}^{1+\log2}\frac{1}{t^2}dt$

I = $\left[\frac{-1}{t}\right]_{1}^{1+\log2}$

I = $\frac{-1}{1+\log2}+1$

I = $\frac{\log2}{1+\log2}$

Therefore, the value of $\int_{1}^{2}\frac{1}{x(1+logx)^2}dx$ is $\frac{\log2}{1+\log2}$ .

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

Solution:

We have,

I = $\int_{0}^{\frac{\pi}{2}}\cos^5xdx$

I = $\int_{0}^{\frac{\pi}{2}}(1-\sin^2x)^2\cos xdx$

Let sin x = t. So, we have

=> cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin x

=> t = sin 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin x

=> t = sin π/2

=> t = 1

So, the equation becomes,

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

I = $\int_{0}^{1}(1+t^4-2t^2)dt$

I = $\left[t-\frac{2}{3}t^3+\frac{t^5}{5}\right]_{0}^{1}$

I = $1-\frac{2}{3}+\frac{1}{5}$

I = $\frac{8}{15}$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\cos^5xdx$ is $\frac{8}{15}$ .

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

Solution:

We have,

I = $\int_{4}^{9}\frac{\sqrt{x}}{(30-x^{\frac{3}{2}})^2}dx$

Let 30 – x^3/2 = t. So, we have

=> $-\frac{3}{2}\sqrt{x}dx$ = dt

=> $\frac{3}{2}\sqrt{x}dx$ = – dt

Now, the lower limit is, x = 4

=> t = 30 – x^3/2

=> t = 30 – 4^3/2

=> t = 30 – 8

=> t = 22

Also, the upper limit is, x = 9

=> t = 30 – x^3/2

=> t = 30 – 9^3/2

=> t = 30 – 27

=> t = 3

So, the equation becomes,

I = $\int_{22}^{3}\frac{-2}{3t^2}dt$

I = $\frac{2}{3}\int_{3}^{22}\frac{1}{t^2}dt$

I = $\frac{2}{3}\left[\frac{-1}{t}\right]_{3}^{22}$

I = $\frac{2}{3}\left[\frac{-1}{22}+\frac{1}{3}\right]$

I = $\frac{2}{3}(\frac{19}{66})$

I = $\frac{19}{99}$

Therefore, the value of $\int_{4}^{9}\frac{\sqrt{x}}{(30-x^{\frac{3}{2}})^2}dx$ is $\frac{19}{99}$ .

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

Solution:

We have,

I = $\int_{0}^{\pi}sin^3x(1+2cosx)(1+cosx)^2dx$

Let cos x = t. So, we have

=> – sin x dx = dt

=> sin x dx = –dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x = π

=> t = cos x

=> t = cos π

=> t = –1

So, the equation becomes,

I = $\int_{0}^{\pi}sin^2x(1+2cosx)(1+cosx)^2.sinxdx$

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

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

I = $\int_{-1}^{1}(1+4t+4t^2-2t^3-5t^4-2t^5)dt$

I = $\left[t+2t^2+\frac{4}{3}t^3-\frac{1}{2}t^4-t^5-\frac{1}{3}t^6\right]_{-1}^{1}$

I = $2+0+\frac{8}{3}-0-2-0$

I = $\frac{8}{3}$

Therefore, the value of $\int_{0}^{\pi}sin^3x(1+2cosx)(1+cosx)^2dx$ is $\frac{8}{3}$ .

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

Solution:

We have,

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

Let sin x = t. So, we have

=> cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin x

=> t = sin 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin x

=> t = sin π/2

=> t = 1

So, the equation becomes,

I = $2\int_{0}^{1}ttan^{-1}tdt$

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

I = $2(\frac{\pi}{4}-\frac{1}{2})$

I = $\frac{\pi}{2}-1$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}2sinxcosxtan^{-1}(sinx)dx$ is $\frac{\pi}{2}-1$ .

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

Solution:

We have,

I = $\int_{0}^{\frac{\pi}{2}}sin2xtan^{-1}(sinx)dx$

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

Let sin x = t. So, we have

=> cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin x

=> t = sin 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin x

=> t = sin π/2

=> t = 1

So, the equation becomes,

I = $2\int_{0}^{1}ttan^{-1}tdt$

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

I = $2(\frac{\pi}{4}-\frac{1}{2})$

I = $\frac{\pi}{2}-1$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}sin2xtan^{-1}(sinx)dx$ is $\frac{\pi}{2}-1$ .

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

Solution:

We have,

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

On using integration by parts, we get,

I = $(cos^{-1}x)^2\int_{0}^{1}dx-\int_0^1(\int dx)\frac{d}{dx}(cos^{-1}x)^2dx$

I = $\left[x(cos^{-1}x)^2\right]_{0}^{1}+2\int_0^1\frac{xcos^{-1}x}{\sqrt{1-x^2}}dx$

Let cos^-1 x = t. So, we have

=> $\frac{-1}{\sqrt{1-x^2}}dx$ = dt

Now, the lower limit is, x = 0

=> t = cos^-1 x

=> t = cos^-1 0

=> t = π/2

Also, the upper limit is, x = 1

=> t = cos^-1 x

=> t = cos^-1 1

=> t = 0

So, the equation becomes,

I = $\left[x(cos^{-1}x)^2\right]_{0}^{1}+2\int_\frac{\pi}{2}^0-tcostdt$

I = $0-0+2\int^\frac{\pi}{2}_0tcostdt$

I = $2\int^\frac{\pi}{2}_0tcostdt$

I = $t\int_{0}^{\frac{\pi}{2}}costdt-\int_0^\frac{\pi}{2}(\int costdt)\frac{dt}{dt}dt$

I = $2\left[tsint-\int sintdt\right]^\frac{\pi}{2}_0$

I = $2\left[tsint+cost\right]^\frac{\pi}{2}_0$

I = $2(\frac{\pi}{2}-1)$

I = π – 2

Therefore, the value of $\int_{0}^{1}(cos^{-1}x)^2dx$ is π – 2.

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

Solution:

We have,

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

Let x = a tan² t. So, we have

=> dx = 2a tan t sec²t dt

Now, the lower limit is, x = 0

=> a tan² t = x

=> a tan² t = 0

=> tan t = 0

=> t = 0

Also, the upper limit is, x = a

=> a tan² t = x

=> a tan² t = a

=> tan² t = 1

=> tan t = 1

=> t = π/4

So, the equation becomes,

I = $\int_{0}^{\frac{\pi}{4}}sin^{-1}\sqrt{\frac{atan^2t}{a+atan^2t}}(2atantsec^2t)dt$

I = $\int_{0}^{\frac{\pi}{4}}sin^{-1}\sqrt{\frac{atan^2t}{a(1+tan^2t)}}(2atantsec^2t)dt$

I = $\int_{0}^{\frac{\pi}{4}}sin^{-1}\sqrt{\frac{tan^2t}{sec^2t}}(2atantsec^2t)dt$

I = $2a\int_{0}^{\frac{\pi}{4}}sin^{-1}(sint)(tantsec^2t)dt$

I = $2a\int_{0}^{\frac{\pi}{4}}t(tantsec^2t)dt$

I = $2a\left[\frac{ttan^2t}{2}-\int \frac{tan^2t}{2}dt\right]_0^\frac{\pi}{4}$

I = $\left[attan^2t-\frac{2a}{2}\int (sec^2t-1)dt\right]_0^\frac{\pi}{4}$

I = $\left[attan^2t-atant+at\right]_0^\frac{\pi}{4}$

I = $\frac{a\pi}{4}tan^2\frac{\pi}{4}-atan\frac{\pi}{4}+a\frac{\pi}{4}$

I = $\frac{a\pi}{4}-a+\frac{a\pi}{4}$

I = $\frac{a\pi}{2}-a$

I = $a(\frac{\pi}{2}-1)$

Therefore, the value of $\int_{0}^{a}sin^{-1}\sqrt{\frac{x}{a+x}}dx$ is $a(\frac{\pi}{2}-1)$ .

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

Solution:

We have,

I = $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sqrt{1+cosx}}{(1-cosx)^{\frac{3}{2}}}dx$

I = $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sqrt{2cos^2\frac{x}{2}}}{(2sin^2\frac{x}{2})^{\frac{3}{2}}}dx$

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

I = $\frac{1}{2}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{cos\frac{x}{2}}{sin^3\frac{x}{2}}dx$

I = $\frac{1}{2}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}cot\frac{x}{2}\cosec^2\frac{x}{2}dx$

Let cot x/2 = t. So, we have

=> $\frac{-1}{2}cosec^2\frac{x}{2}dx$ = dt

Now, the lower limit is, x = π/3

=> t = cot x/2

=> t = cot π/6

=> t = √3

Also, the upper limit is, x = π/2

=> t = cot x/2

=> t = cot π/4

=> t = 1

So, the equation becomes,

I = $-\int_{\sqrt{3}}^{1}tdt$

I = $\int^{\sqrt{3}}_{1}tdt$

I = $\left[\frac{t^2}{2}\right]^{\sqrt{3}}_{1}$

I = $\frac{3}{2}-\frac{1}{2}$

I = 1

Therefore, the value of $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sqrt{1+cosx}}{(1-cosx)^{\frac{3}{2}}}dx$ is 1.

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

Solution:

We have,

I = $\int_{0}^{a}x\sqrt{\frac{a^2-x^2}{a^2+x^2}}dx$

Let x² = a² cos 2t. So, we have

=> 2x dx = – 2a² sin 2t dt

Now, the lower limit is, x = 0

=> a² cos 2t = x²

=> a² cos 2t = 0

=> cos 2t = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is, x = a

=> a² cos 2t = x²

=> a² cos 2t = a²

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

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

I = $-a^2\int_{\frac{\pi}{4}}^{0}\frac{sint}{cost}sin2tdt$

I = $a^2\int^{\frac{\pi}{4}}_{0}\frac{sint}{cost}sin2tdt$

I = $a^2\int^{\frac{\pi}{4}}_{0}2sin^2tdt$

I = $a^2\int^{\frac{\pi}{4}}_{0}(1-cos2t)dt$

I = $a^2\left[t-\frac{sin2t}{2}\right]^{\frac{\pi}{4}}_{0}$

I = $a^2(\frac{\pi}{4}-\frac{1}{2})$

Therefore, the value of $\int_{0}^{a}x\sqrt{\frac{a^2-x^2}{a^2+x^2}}dx$ is $a^2(\frac{\pi}{4}-\frac{1}{2})$ .

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

Solution:

We have,

I = $\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}dx$

Let x = a cos 2t. So, we have

=> dx = –2a sin 2t

Now, the lower limit is, x = –a

=> a cos 2t = x

=> a cos 2t = –a

=> cos 2t = –1

=> 2t = π

=> t = π/2

Also, the upper limit is, x = a

=> a cos 2t = x

=> a cos 2t = a

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I = $\int_{\frac{\pi}{2}}^{0}\sqrt{\frac{a-acos2t}{a+acos2t}}(-2asin2t)dt$

I = $-2a\int_{\frac{\pi}{2}}^{0}\sqrt{\frac{a(1-cos2t)}{a(1+cos2t)}}sin2tdt$

I = $-2a\int_{\frac{\pi}{2}}^{0}\sqrt{\frac{1-cos2t}{1+cos2t}}sin2tdt$

I = $-2a\int_{\frac{\pi}{2}}^{0}\sqrt{\frac{2sin^2t}{2cos^2t}}sin2tdt$

I = $-2a\int_{\frac{\pi}{2}}^{0}\frac{sint}{cost}(2sintcost)dt$

I = $-4a\int_{\frac{\pi}{2}}^{0}sin^2tdt$

I = $\frac{4a}{2}\int^{\frac{\pi}{2}}_{0}(1-cos2t)dt$

I = $2a\int^{\frac{\pi}{2}}_{0}(1-cos2t)dt$

I = $2a\left[t-\frac{sin2t}{2}\right]^{\frac{\pi}{2}}_{0}$

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

I = πa

Therefore, the value of $\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}dx$ is πa.

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

Solution:

We have,

I = $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{cos^2x+3cosx+2}dx$

Let cos x = t. So, we have

=> – sin x dx = dt

=> sin x dx = –dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x = π/2

=> t = cos x

=> t = cos π/2

=> t = 0

So, the equation becomes,

I = $\int_{1}^{0}\frac{-t}{t^2+3t+2}dt$

I = $\int_{0}^{1}\frac{t}{(t+2)(t+1)}dt$

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

I = $\left[-log|1+t|+2log|t+2|\right]_{0}^{1}$

I = – log 2 + 2 log 3 + 0 – 2 log 2

I = 2 log 3 – 3 log 2

I = log 9 – log 8

I = log 9/8

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{cos^2x+3cosx+2}dx$ is log 9/8.

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

Solution:

We have,

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

I = $\int_{0}^{\frac{\pi}{2}}\frac{\frac{sinx}{cosx}}{1+m^2(\frac{sin^2x}{cos^2x})}dx$

I = $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{cos^2x+m^2sin^2x}dx$

Let sin² x = t. So, we have

=> 2 sin x cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin² x

=> t = sin² 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin² x

=> t = sin² π/2

=> t = 1

So, the equation becomes,

I = $\frac{1}{2}\int_{0}^{1}\frac{1}{(1-t)+m^2t}dt$

I = $\frac{1}{2}\int_{0}^{1}\frac{1}{(m^2-1)t+1}dt$

I = $\frac{1}{2(m^2-1)}\left[\log|(m^2-1)t+1|\right]_0^1$

I = $\frac{1}{2(m^2-1)}\left[\log|m^2-1+1|-log1\right]$

I = $\frac{1}{2(m^2-1)}\left[\log m^2-log1\right]$

I = $\frac{\log m^2}{2(m^2-1)}$

I = $\frac{2\log m}{2(m^2-1)}$

I = $\frac{\log m}{m^2-1}$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{tanx}{1+m^2tan^2x}dx$ is $\frac{\log m}{m^2-1}$ .

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

Solution:

We have,

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

Let x = sin t. So, we have

=> dx = cos t dt

Now, the lower limit is, x = 0

=> sin t = x

=> sin t = 0

=> t = 0

Also, the upper limit is, x = 1/2

=> sin t = x

=> sin t = 1/2

=> t = π/6

So, the equation becomes,

I = $\int_{0}^{\frac{\pi}{6}}\frac{1}{(1+sin^2t)(\sqrt{1-sin^2t})}(cost)dt$

I = $\int_{0}^{\frac{\pi}{6}}\frac{1}{(1+sin^2t)(\sqrt{cos^2t})}(cost)dt$

I = $\int_{0}^{\frac{\pi}{6}}\frac{1}{(1+sin^2t)cost}(cost)dt$

I = $\int_{0}^{\frac{\pi}{6}}\frac{1}{1+sin^2t}dt$

I = $\int_{0}^{\frac{\pi}{6}}\frac{sec^2t}{1+2tan^2t}dt$

Let tan t = u. So, we have

=> sec² t dt = du

Now, the lower limit is, t = 0

=> u = tan t

=> u = tan 0

=> t = 0

Also, the upper limit is, t = π/6

=> u = tan t

=> u = tan π/6

=> t = 1/√3

So, the equation becomes,

I = $\int_{0}^{\frac{1}{\sqrt{3}}}\frac{1}{1+2u^2}du$

I = $\frac{1}{\sqrt{2}}\left[tan^{-1}(\sqrt{2}u)\right]_{0}^{\frac{1}{\sqrt{3}}}$

I = $\frac{1}{\sqrt{2}}\tan^{-1}\sqrt{\frac{2}{3}}$

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

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

Solution:

We have,

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

Let 1/x² – 1 = t. So, we have

=> –2/x³ dx = dt

Now, the lower limit is, x = 1/3

=> t = 1/x² – 1

=> t = 9 – 1

=> t = 8

Also, the upper limit is, x = 1

=> t = 1/x² – 1

=> t = 1 – 1

=> t = 0

So, the equation becomes,

I = $\frac{-1}{2}\int_{8}^{0}t^{\frac{1}{3}}dt$

I = $\frac{1}{2}\left[\frac{t^{\frac{4}{3}}}{\frac{4}{3}}\right]_{0}^{8}$

I = $\frac{3}{8}\left[t^{\frac{4}{3}}\right]_{0}^{8}$

I = $\frac{3}{8}(8^{\frac{4}{3}})$

I = $\frac{3}{8}(16)$

I = 6

Therefore, the value of $\int_{\frac{1}{3}}^{1}\frac{(x-x^3)^{\frac{1}{3}}}{x^4}dx$ is 6.

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

Solution:

We have,

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

I = $\int_{0}^{\frac{\pi}{4}}\frac{tan^2xsec^2x}{tan^6x+2tan^3x+1}dx$

Let tan x = t. So, we have

=> sec² x dx = dt

Now, the lower limit is, x = 0

=> t = tan t

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/4

=> t = tan t

=> t = tan π/4

=> t = 1

So, the equation becomes,

I = $\int_{0}^{1}\frac{t^2}{t^6+2t^3+1}dt$

Let t³= u. So, we have

=> 3t² dt = du

=> t² dt = du/3

Now, the lower limit is, t = 0

=> u = t³

=> u = 0³

=> u = 0

Also, the upper limit is, t = 1

=> u = t³

=> u = 1³

=> u = 1

So, the equation becomes,

I = $\frac{1}{3}\int_{0}^{1}\frac{1}{u^2+2u+1}du$

I = $\frac{1}{3}\int_{0}^{1}\frac{1}{(u+1)^2}du$

I = $\frac{1}{3}\left[\frac{-1}{u+1}\right]_{0}^{1}$

I = $\frac{1}{3}\left[\frac{-1}{1+1}+\frac{1}{1+0}\right]$

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

I = $\frac{1}{3}(\frac{1}{2})$

I = $\frac{1}{6}$

Therefore, the value of $\int_{0}^{\frac{\pi}{4}}\frac{sin^2xcos^2x}{(sin^3x+cos^3x)^2}dx$ is $\frac{1}{6}$ .

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

Solution:

We have,

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

I = $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx-cos^3x}tan^2xcos^2xdx$

I = $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx-cos^3x}sin^2xdx$

I = $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx(1-cos^2x)}sin^2xdx$

I = $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx}sin^3xdx$

Let cos x = t. So, we have

=> – sin x dx = dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x = π/2

=> t = cos x

=> t = cos π/2

=> t = 0

So, the equation becomes,

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

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

I = $\int_{0}^{1}(t^{\frac{1}{2}}-t^{\frac{5}{2}})dt$

I = $\left[\frac{2t^{\frac{3}{2}}}{3}-\frac{2t^{\frac{7}{2}}}{7}\right]_{0}^{1}$

I = $\frac{2}{3}-\frac{2}{7}$

I = $\frac{8}{21}$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\sqrt{cosx-cos^3x}(sec^2x-1)cos^2xdx$ is $\frac{8}{21}$ .

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

Solution:

We have,

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

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

I = $\int_{0}^{\frac{\pi}{2}}\frac{cos\frac{x}{2}-sin\frac{x}{2}}{(cos\frac{x}{2}+sin\frac{x}{2})^{n-1}}dx$

Let cos x/2 + sin x/2 = t. So, we have

=> (cos x/2 – sin x/2) dx = 2 dt

Now, the lower limit is, x = 0

=> t = cos x/2 + sin x/2

=> t = cos 0 + sin 0

=> t = 1 + 0

=> t = 1

Also, the upper limit is, x = π/2

=> t = cos x/2 + sin x/2

=> t = cos π/2 + sin π/2

=> t = 1/√2 + 1/√2

=> t = √2

So, the equation becomes,

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

I = $\left[\frac{2t^{-n+2}}{-n+2}\right]_{1}^{\sqrt{2}}$

I = $\frac{2}{2-n}\left[(\sqrt{2})^{2-n}-1\right]$

I = $\frac{2}{2-n}\left[2^{1-\frac{n}{2}}-1\right]$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{cosx}{(cos\frac{x}{2}+sin\frac{x}{2})^n}dx$ is $\frac{2}{2-n}\left[2^{1-\frac{n}{2}}-1\right]$ .

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Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.2 | Set 2

Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.3 | Set 1

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Chapter 1: Relations

Chapter 2: Functions

Chapter 3: Binary Operations

Chapter 4: Inverse Trigonometric Functions

Chapter 5: Algebra of Matrices

Chapter 6: Determinants

Chapter 7: Adjoint and Inverse of a Matrix

Chapter 8: Solutions of Simultaneous Linear Equations

Chapter 9: Continuity

Chapter 10: Differentiability

Chapter 11: Differentiation

Chapter 12: Higher Order Derivatives

Chapter 13: Derivative as a Rate Measurer

Chapter 14: Differentials, Errors, and Approximations

Chapter 15: Mean Value Theorems

Chapter 16: Tangents and Normals

Chapter 17: Increasing and Decreasing Functions

Chapter 18: Maxima and Minima

Chapter 19: Indefinite Integrals

Chapter 20: Definite Integrals

Chapter 21: Areas of Bounded Regions

Chapter 22: Differential Equations

Chapter 23: Algebra of Vectors

Chapter 24: Scalar Or Dot Product

Chapter 25: Vector or Cross Product

Chapter 26: Scalar Triple Product

Chapter 27: Direction Cosines and Direction Ratios

Chapter 28: Straight Line in Space

Chapter 29: The Plane

Chapter 30: Linear programming

Chapter 31: Probability

Chapter 32: Mean and Variance of a Random Variable

Chapter 33: Binomial Distribution

Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.2 | Set 3

Evaluate the following definite integrals:

Question 42.

Question 43.

Question 44.

Question 45.

Question 46.

Question 47.

Question 48.

Question 49.

Question 50.

Question 51.

Question 52.

Question 53.

Question 54.

Question 55.

Question 56.

Question 57.

Question 58.

Question 59.

Question 60.

Question 61.

Question 62.

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