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

Last Updated : 30 Jun, 2021

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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

=> sec²x 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

=> sec²x 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 $\frac{1}{8}$ .

My Personal Notes

Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.2 | Set 1

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

1. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.5 | Set 3

2. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.5 | Set 1

3. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.5 | Set 2

4. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.1 | Set 1

5. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.1 | Set 2

6. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.1 | Set 3

7. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.2 | Set 1

8. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.2 | Set 3

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

10. Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.3 | Set 2

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

Evaluate the following definite integrals:

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

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

Question 23. $\int_{0}^{1}tan^{-1}xdx$

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

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

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

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

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

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

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

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

Question 32. $\int_{0}^{1}xtan^{-1}xdx$

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

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

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

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

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

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

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

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

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

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Related Articles

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

Evaluate the following definite integrals:

Question 21.

Question 22.

Question 23.

Question 24.

Question 25.

Question 26.

Question 27.

Question 28.

Question 29.

Question 30.

Question 31.

Question 32.

Question 33.

Question 34.

Question 35.

Question 36.

Question 37.

Question 38.

Question 39.

Question 40.

Question 41.

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

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

Question 23. $\int_{0}^{1}tan^{-1}xdx$

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

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

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

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

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

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

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

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

Question 32. $\int_{0}^{1}xtan^{-1}xdx$

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

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

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

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

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

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

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

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

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