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

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

Last Updated : 30 Jun, 2021

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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