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

Evaluate the following definite integrals:

Question 21. \int_{0}^{\pi}\frac{sinx}{sinx+cosx}dx

Solution:

We have,

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

Let sin x = A (sin x + cos x) + B\frac{d}{dx}(sinx+cosx)

=> sin x = A (sin x + cos x) + B (cos x – sin x)

=> sin x = sin x (A – B) + cos x (A + B)

On comparing both sides, we get

A – B = 1 and A + B = 0

On solving, we get A = 1/2 and B = –1/2.

Therefore, the expression becomes,

I = \frac{1}{2}\int_0^\pi dx-\frac{1}{2}\int_{0}^{\pi}\frac{cosx-sinx}{sinx+cosx}dx

I = \left[\frac{x}{2}\right]_0^\pi-\frac{1}{2}\left[\log(sinx+cosx)\right]_{0}^{\pi}

I = \frac{\pi}{2}-\frac{1}{2}(0)

I = \frac{\pi}{2}

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

Question 22. \int_{0}^{\pi}\frac{1}{3+2sinx+cosx}dx

Solution:

We have,

I = \int_{0}^{\pi}\frac{1}{3+2sinx+cosx}dx

On putting cos x = \frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}  and sin x = \frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}      , we get,

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

I = \int_{0}^{\pi}\frac{1+tan^2\frac{x}{2}}{3(1+tan^2\frac{x}{2})+4tan\frac{x}{2}+1-tan^2\frac{x}{2}}dx

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

I = \int_{0}^{\pi}\frac{sec^2\frac{x}{2}}{2tan^2\frac{x}{2}+4tan\frac{x}{2}+4}dx

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

=> 1/2 sec² x/2 dx = dt

=> sec² x/2 dx = 2 dt

Now, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π

=> t = tan x/2

=> t = tan π/2

=> t = ∞

So, the equation becomes,

I = \int_{0}^{\infty}\frac{2}{2t^2+4t+4}dt

I = \int_{0}^{\infty}\frac{1}{t^2+2t+2}dt

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

I = \left[tan^{-1}(t+1)\right]^\infty_0

I = tan^{-1}\infty-tan^{-1}1

I = \frac{\pi}{2}-\frac{\pi}{4}

I = \frac{\pi}{4}

Therefore, the value of \int_{0}^{\pi}\frac{1}{3+2sinx+cosx}dx  is \frac{\pi}{4} .

Question 23. \int_{0}^{1}tan^{-1}xdx

Solution:

We have,

I = \int_{0}^{1}tan^{-1}xdx

I = tan^{-1}x\int_{0}^{1}dx-\int_{0}^1(\int dx) \frac{d}{dx}(tan^{-1}x)dx

I = \left[xtan^{-1}x\right]_{0}^{1}-\int_{0}^1\frac{x}{1+x^2}dx

I = \left[xtan^{-1}x\right]_{0}^{1}-\frac{1}{2}\int_{0}^1\frac{2x}{1+x^2}dx

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

I = (\frac{\pi}{4}-0)-\frac{1}{2}\left[log(1+1)-log(1+0)\right]

I = \frac{\pi}{4}-\frac{1}{2}\left[log2-log1\right]

I = \frac{\pi}{4}-\frac{1}{2}\left[log2-0\right]

I = \frac{\pi}{4}-\frac{\log2}{2}

Therefore, the value of \int_{0}^{1}tan^{-1}xdx  is \frac{\pi}{4}-\frac{\log2}{2} .

Question 24. \int_{0}^{\frac{1}{2}}\frac{xsin^{-1}x}{\sqrt{1-x^2}}dx

Solution:

We have,

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

Let sin^–1 x = t. So, we have

=> \frac{1}{\sqrt{1+x^2}}dx  = dt

Now, the lower limit is, x = 0

=> t = sin^–1 x

=> t = sin^–1 0

=> t = 0

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

=> t = sin^–1 x

=> t = sin^–1 1/2

=> t = π/6

So, the equation becomes,

I = \int_{0}^{\frac{\pi}{6}}tsintdt

I = t\int_{0}^{\frac{\pi}{6}}sintdt-\int_{0}^{\frac{\pi}{6}}(\int sintdt) \frac{d}{dt}(t)dt

I = \left[-tcost\right]_{0}^{\frac{\pi}{6}}+\int_{0}^{\frac{\pi}{6}}costdt

I = \left[-tcost\right]_{0}^{\frac{\pi}{6}}+\left[sint\right]_{0}^{\frac{\pi}{6}}

I = \left[-\frac{\pi}{6}(\frac{\sqrt{3}}{2})+0\right]+\left[sin\frac{\pi}{6}-0\right]

I = \frac{-\sqrt{3}\pi}{12}+\frac{1}{2}

I = \frac{1}{2}-\frac{\sqrt{3}\pi}{12}

Therefore, the value of \int_{0}^{\frac{1}{2}}\frac{xsin^{-1}x}{\sqrt{1-x^2}}dx  is \frac{1}{2}-\frac{\sqrt{3}\pi}{12} .

Question 25. \int_{0}^{\frac{\pi}{4}}(\sqrt{tanx}+\sqrt{cotx})dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{4}}(\sqrt{tanx}+\sqrt{cotx})dx

I = \int_{0}^{\frac{\pi}{4}}(\sqrt{\frac{sinx}{cosx}}+\sqrt{\frac{cosx}{sinx}})dx

I = \int_{0}^{\frac{\pi}{4}}(\frac{sinx+cosx}{\sqrt{sinxcosx}})dx

I = \sqrt{2}\int_{0}^{\frac{\pi}{4}}(\frac{sinx+cosx}{\sqrt{2sinxcosx}})dx

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

Let sinx – cosx = t. So, we have

=> (cos x + sin x) dx = dt

Now, the lower limit is, x = 0

=> t = sinx – cosx

=> t = sin 0 – cos 0

=> t = 0 – 1

=> t = –1

Also, the upper limit is, x = π/4

=> t = sinx – cosx

=> t = sin π/4 – cos π/4

=> t = sin π/4 – sin π/4

=> t = 0

So, the equation becomes,

I = \sqrt{2}\int_{-1}^{0}\frac{1}{\sqrt{1-t^2}}dt

I = \sqrt{2}\left[sin^{-1}t\right]_{-1}^{0}

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

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

I = \sqrt{2}(\frac{\pi}{2})

I = \frac{\pi}{\sqrt{2}}

Therefore, the value of \int_{0}^{\frac{\pi}{4}}(\sqrt{tanx}+\sqrt{cotx})dx  is \frac{\pi}{\sqrt{2}} .

Question 26. \int_{0}^{\frac{\pi}{4}}\frac{tan^3x}{1+cos2x}dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{4}}\frac{tan^3x}{1+cos2x}dx

I = \int_{0}^{\frac{\pi}{4}}\frac{tan^3x}{2cos^2x}dx

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

I = \frac{1}{2}\int_{0}^{\frac{\pi}{4}}tan^3xsec^2xdx

Let tan x = t. So, we have

=> sec² x dx = dt

Now, the lower limit is, x = 0

=> t = tan x

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/4

=> t = tan x

=> t = tan π/4

=> t = 1

So, the equation becomes,

I = \frac{1}{2}\int_{0}^{1}t^3dt

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

I = \frac{1}{2}(\frac{1}{4}-0)

I = \frac{1}{8}

Therefore, the value of \int_{0}^{\frac{\pi}{4}}\frac{tan^3x}{1+cos2x}dx  is \frac{1}{8} .

Question 27. \int_{0}^{\pi}\frac{1}{5+3cosx}dx

Solution:

We have,

I = \int_{0}^{\pi}\frac{1}{5+3cosx}dx

On putting cos x = \frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}} , we get

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

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

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

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

=> 1/2 sec² x/2 dx = dt

=> sec² x/2 dx = 2 dt

Now, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π

=> t = tan x/2

=> t = tan π/2

=> t = ∞

So, the equation becomes,

I = \int_{0}^{\infty}\frac{1}{5(1+t^2)+3(1-t^2)}dt

I = \int_{0}^{\infty}\frac{1}{5+5t^2+3-3t^2}dt

I = \int_{0}^{\infty}\frac{1}{8+2t^2}dt

I = \frac{1}{2}\int_{0}^{\infty}\frac{1}{4+t^2}dt

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

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

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

I = \frac{1}{2}(\frac{\pi}{2})

I = \frac{\pi}{4}

Therefore, the value of \int_{0}^{\pi}\frac{1}{5+3cosx}dx  is \frac{\pi}{4} .

Question 28. \int_{0}^{\frac{\pi}{2}}\frac{1}{a^2sin^2x+b^2cos^2x}dx

Solution:

We have,

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

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

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

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

Let tan x = t. So, we have

=> sec² x dx = dt

Now, the lower limit is, x = 0

=> t = tan x

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x

=> t = tan π/2

=> t = ∞

So, the equation becomes,

I = \frac{1}{a^2}\int_{0}^{\infty}\frac{1}{t^2+\frac{b^2}{a^2}}dt

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

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

I = \frac{1}{a^2}(\frac{a}{b})(\frac{\pi}{2})

I = \frac{\pi}{2ab}

Therefore, the value of \int_{0}^{\frac{\pi}{2}}\frac{1}{a^2sin^2x+b^2cos^2x}dx  is \frac{\pi}{2ab} .

Question 29. \int_{0}^{\frac{\pi}{2}}\frac{x+sinx}{1+cosx}dx

Solution:

We have,

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

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

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

I = \left[xtan\frac{x}{2}-\int_{0}^{\frac{\pi}{2}}tan\frac{x}{2}dx+\int_{0}^{\frac{\pi}{2}}tan\frac{x}{2}dx\right]^{\frac{\pi}{2}}_0

I = \left[xtan\frac{x}{2}\right]^\frac{\pi}{2}_0

I = \frac{\pi}{2}tan\frac{\pi}{4}-0

I = \frac{\pi}{2}

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

Question 30. \int_{0}^{1}\frac{tan^{-1}x}{1+x^2}dx

Solution:

We have,

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

Let tan^–1 x = t. So, we have

=> \frac{1}{1+x^2}dx  = dt

Now, the lower limit is, x = 0

=> t = tan^–1 x

=> t = tan^–1 0

=> t = 0

Also, the upper limit is, x = 1

=> t = tan^–1 x

=> t = tan^–1 1

=> t = π/4

So, the equation becomes,

I = \int_{0}^{\frac{\pi}{4}}tdt

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

I = \frac{1}{2}(\frac{\pi^2}{16})-0

I = \frac{\pi^2}{32}

Therefore, the value of \int_{0}^{1}\frac{tan^{-1}x}{1+x^2}dx  is \frac{\pi^2}{32} .

Question 31. \int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{3+sin2x}dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{3+sin2x}dx

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

I = \int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{4-(cosx-sinx)^2}dx

I = \left[\frac{1}{4}log|\frac{2+sinx-cosx}{2-sinx+cosx}|\right]_{0}^{\frac{\pi}{4}}

I = \frac{1}{4}\left[log|\frac{2+sin\frac{\pi}{4}-cos\frac{\pi}{4}}{2-sin\frac{\pi}{4}+cos\frac{\pi}{4}}|-log\frac{2+0-1}{2-0+1}\right]

I = \frac{1}{4}\left(log|\frac{2}{2}|-log\frac{1}{3}\right)

I = \frac{1}{4}\left(log1-log\frac{1}{3}\right)

I = \frac{1}{4}\left(-log\frac{1}{3}\right)

I = -\frac{log\frac{1}{3}}{4}

I = \frac{log3}{4}

Therefore, the value of \int_{0}^{\frac{\pi}{4}}\frac{sinx+cosx}{3+sin2x}dx  is \frac{log3}{4} .

Question 32. \int_{0}^{1}xtan^{-1}xdx

Solution:

We have,

I = \int_{0}^{1}xtan^{-1}xdx

On using integration by parts, we get

I = tan^{-1}x\int_{0}^{1}xdx-\int_0^1(\int xdx)\frac{d}{dx}(tan^{-1}x)dx

I = \left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_0^1\frac{x^2}{1+x^2}dx

I = \left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_0^1\frac{1+x^2-1}{1+x^2}dx

I = \left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_0^1\frac{1+x^2}{1+x^2}dx+\frac{1}{2}\int_0^1\frac{1}{1+x^2}dx

I = \left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_0^1\frac{1+x^2}{1+x^2}dx+\frac{1}{2}\int_0^1\frac{1}{1+x^2}dx

I = \left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_0^1dx+\frac{1}{2}\int_0^1\frac{1}{1+x^2}dx

I = \left[\frac{x^2tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\left[x\right]_0^1+\frac{1}{2}\left[tan^{-1}x\right]_0^1

I = (\frac{\pi}{8}-0)-\frac{1}{2}(1-0)+\frac{1}{2}\left[tan^{-1}1-tan^{-1}0\right]

I = \frac{\pi}{8}-\frac{1}{2}+\frac{1}{2}(\frac{\pi}{4}-0)

I = \frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}

I = \frac{\pi}{4}-\frac{1}{2}

Therefore, the value of \int_{0}^{1}xtan^{-1}xdx  is \frac{\pi}{4}-\frac{1}{2} .

Question 33. \int_{0}^{1}\frac{1-x^2}{x^4+x^2+1}dx

Solution:

We have,

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

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

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

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

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

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

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

I = -\frac{1}{2}\log|\frac{1^2-1+1}{1^2+1+1}|+\frac{1}{2}\log|\frac{0^2-0+1}{0^2+0+1}|

I = -\frac{1}{2}\log\frac{1}{3}+\frac{1}{2}\log1

I = -\frac{1}{2}\log\frac{1}{3}+\frac{1}{2}(0)

I = -\frac{1}{2}\log\frac{1}{3}

I = \log(\frac{1}{3})^{\frac{-1}{2}}

I = \log3^{\frac{1}{2}}

I = \frac{1}{2}\log3

Therefore, the value of \int_{0}^{1}\frac{1-x^2}{x^4+x^2+1}dx  is \frac{1}{2}\log3 .

Question 34. \int_{0}^{1}\frac{24x^3}{(1+x^2)^4}dx

Solution:

We have,

I = \int_{0}^{1}\frac{24x^3}{(1+x^2)^4}dx

Let 1 + x² = t. So, we have

=> 2x dx = dt

Now, the lower limit is, x = 0

=> t = 1 + x²

=> t = 1 + 0²

=> t = 1 + 0

=> t = 1

Also, the upper limit is, x = π

=> t = 1 + x²

=> t = 1 + 1²

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I = \int_{0}^{2}\frac{12(2x)(x^2)}{(1+x^2)^4}dx

I = \int_{1}^{2}\frac{12(t-1)}{t^4}dt

I = 12\int_{1}^{2}(\frac{t}{t^4}-\frac{1}{t^4})dt

I = 12\int_{1}^{2}(\frac{1}{t^3}-\frac{1}{t^4})dt

I = 12\left[\frac{-1}{2t^2}-\frac{1}{3t^3}\right]_{1}^{2}

I = 12\left[\frac{-1}{8}+\frac{1}{24}+\frac{1}{2}-\frac{1}{3}\right]

I = 12\left[\frac{-3+1+12-8}{24}\right]

I = 12(\frac{2}{24})

I = 1

Therefore, the value of \int_{0}^{1}\frac{24x^3}{(1+x^2)^4}dx  is 1.

Question 35. \int_{4}^{12}x(x-4)^{\frac{1}{3}}dx

Solution:

We have,

I = \int_{4}^{12}x(x-4)^{\frac{1}{3}}dx

Let x – 4 = t³. So, we have

=> dx = 3t² dt

Now, the lower limit is, x = 4

=> t³ = x – 4

=> t³ = 4 – 4

=> t³ = 0

=> t = 0

Also, the upper limit is, x = 12

=> t³ = x – 4

=> t³ = 12 – 4

=> t³ = 8

=> t = 2

So, the equation becomes,

I = \int_{0}^{2}(t^3+4)t(3t^2)dt

I = \int_{0}^{2}3t^3(t^3+4)dt

I = 3\int_{0}^{2}(t^6+4t^3)dt

I = 3\left[\frac{t^7}{7}+t^4\right]_{0}^{2}

I = 3\left[\frac{128}{7}+16-0-0\right]

I = \frac{720}{7}

Therefore, the value of \int_{4}^{12}x(x-4)^{\frac{1}{3}}dx  is \frac{720}{7} .

Question 36. \int_{0}^{\frac{\pi}{2}}x^2sinxdx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}}x^2sinxdx

On using integration by parts, we get

I = x^2\int sinxdx-\int(\int sinxdx)\frac{d}{dx}(x^2)dx

I = x^2cosx-\int 2xcosxdx

I = x^2cosx+2[x\int cosxdx-\int(\int cosxdx)\frac{dx}{dx}dx]

I = x^2cosx+2[xsinxdx-\int sinxdx]

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

I = π + 0 – 0 – 0 – 2

I = π – 2

Therefore, the value of \int_{0}^{\frac{\pi}{2}}x^2sinxdx  is π – 2.

Question 37. \int_{0}^{1}\sqrt{\frac{1-x}{1+x}}dx

Solution:

We have,

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

Let x = cos 2t. So, we have

=> dx = – 2 sin 2t dt

Now, the lower limit is, x = 0

=> cos 2t = x

=> cos 2t = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is, x = 1

=> cos 2t = x

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

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

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

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

I = 4\int^{\frac{\pi}{4}}_{0}sin^2tdt

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

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

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

I = 2\left[\frac{\pi}{4}-\frac{1}{2}\right]

I = \frac{\pi}{2}-1

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

Question 38. \int_{0}^{1}\frac{1-x^2}{(1+x^2)^2}dx

Solution:

We have,

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

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

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

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

=> (1 – 1/x²)dx = dt

Now, the lower limit is, x = 0

=> t = x + 1/x

=> t = ∞

Also, the upper limit is, x = 1

=> t = x + 1/x

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I = -\int_{\infty}^{2}\frac{dt}{t^2}

I = \int^{\infty}_{2}\frac{dt}{t^2}

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

I = \frac{-1}{\infty}+\frac{1}{2}

I = \frac{1}{2}

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

Question 39. \int_{-1}^{1}5x^4\sqrt{x^2+1}dx

Solution:

We have,

I = \int_{-1}^{1}5x^4\sqrt{x^2+1}dx

Let x⁵ + 1 = t. So, we have

=> 5x⁴ dx = dt

Now, the lower limit is, x = –1

=> t = x⁵ + 1

=> t = (–1)⁵ + 1

=> t = –1 + 1

=> t = 0

Also, the upper limit is, x = 1

=> t = x⁵ + 1

=> t = (1)⁵ + 1

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I = \int_{0}^{2}\sqrt{t}dt

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

I = \frac{2}{3}(2^{\frac{3}{2}})

I = \frac{2}{3}(2\sqrt{2})

I = \frac{4\sqrt{2}}{3}

Therefore, the value of \int_{-1}^{1}5x^4\sqrt{x^2+1}dx  is \frac{4\sqrt{2}}{3} .

Question 40. \int_{0}^{\frac{\pi}{2}}\frac{cos^2x}{1+3sin^2x}dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}}\frac{cos^2x}{1+3sin^2x}dx

I = \int_{0}^{\frac{\pi}{2}}\frac{sec^2x}{sec^2x(sec^2x+3tan^2x)}dx

Let tan x = t. So, we have

=> sec² x dx = dt

Now, the lower limit is, x = 0

=> t = tan x

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x

=> t = tan π/2

=> t = ∞

So, the equation becomes,

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

I = \frac{-1}{3}\int_{0}^{\infty}(\frac{1}{1+t^2}-\frac{1}{1+4t^2})dt

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

I = \frac{-1}{3}(\frac{\pi}{2})

I = \frac{-\pi}{6}

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

Question 41. \int_{0}^{\frac{\pi}{4}}sin^32tcos2tdt

Solution:

We have,

I = \int_{0}^{\frac{\pi}{4}}sin^32tcos2tdt

Let sin 2t = u. So, we have

=> 2 cos 2t dt = du

=> cos 2t dt = du/2

Now, the lower limit is, x = 0

=> u = sin 2t

=> u = sin 0

=> u = 0

Also, the upper limit is, x = π/4

=> u = sin 2t

=> u = sin π/2

=> u = 1

So, the equation becomes,

I = \frac{1}{2}\int_{0}^{1}u^3du

I = \frac{1}{2}\left[\frac{u^4}{4}\right]_{0}^{1}

I = \frac{1}{2}(\frac{1}{4})

I = \frac{1}{8}

Therefore, the value of \int_{0}^{\frac{\pi}{4}}sin^32tcos2tdt  is \frac{1}{8} .

Summary

This exercise typically covers:

• More complex definite integrals involving trigonometric functions
• Integrals with algebraic and exponential functions
• Application of integration techniques like substitution and integration by parts
• Evaluation of definite integrals using symmetry and periodicity
• Integrals involving composite functions