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 sec² t 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] .

Summary

This exercise typically focuses on:

• Advanced techniques for evaluating definite integrals
• Integration involving trigonometric functions
• Definite integrals with rational functions
• Application of substitution method in definite integrals
• Evaluation of definite integrals using symmetry properties