Class 12 RD Sharma Solutions- Chapter 20 Definite Integrals - Exercise 20.4 Part A

Last Updated : 2 Sep, 2024

Evaluate each of the following integrals (1-16):

Question 1. \int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx

Solution:

We know that \int\limits_0^{2π }f(x)dx=\int\limits_0^{2π }f(2π -x)dx

so,

\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx =\int\limits_0^{2π}\frac{e^{sin(2π-x)}}{e^{sin(2π-x)}+e^{-sin(2π-x)}}dx

we know, sin(2π -x)=-sinx

\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx =\int\limits_0^{2π}\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}}dx

if 

I = \int\limits_0^{2π}\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}}dx

then 

I =\int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx

2I = \int\limits_0^{2π}\frac{e^{sinx}}{e^{sinx}+e^{-sinx}}dx +\int\limits_0^{2π}\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}}dx

2I=\int\limits_0^{2π}(\frac{e^{sinx}}{e^{sinx}+e^{-sinx}} +\frac{e^{-sinx}}{e^{-sinx}+e^{sinx}})dx

2I=\int\limits_0^{2π}(\frac{e^{sinx}+e^{-sinx}}{e^{sinx}+e^{-sinx}} )dx

2I=\int\limits_0^{2π}dx

2I=2π 

l=π 

Question 2. \int\limits_0^{2π} log (secx+tanx)dx

Solution:

We know that \int\limits_0^{2π }f(x)dx=\int\limits_0^{2π }f(2π -x)dx

So, 

\int\limits_0^{2π} log (secx+tanx)dx= \int\limits_0^{2π} log (sec(2π-x)+tan(2π-x)dx

\int\limits_0^{2π} log (secx+tanx)dx= \int\limits_0^{2π} log (secx-tanx)dx

if 

I = \int\limits_0^{2π} log (secx-tanx)dx

then 

I = \int\limits_0^{2π} log (secx+tanx)dx

2I = \int\limits_0^{2π} log (secx+tanx)dx+ \int\limits_0^{2π} log (secx-tanx)dx

2I= \int\limits_0^{2π} log (sec^2x-tan^2x)dx

2I = \int\limits_0^{2π} log (1)dx

2I=0

I=0

Question 3. \int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx

Solution:

We know \int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx

So,

\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx }+\sqrt{cotx}}dx = \int\limits_{π/6}^{π/3}\frac{\sqrt{tan(π/2-x)}}{\sqrt{tan(π/2-x)}+\sqrt{cot(π/2-x)}}dx

\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx }+\sqrt{cotx}}dx = \int\limits_{π/6}^{π/3}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx

if

I=\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx

then 

I=\int\limits_{π/6}^{π/3}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx

So

2I=\int\limits_{π/6}^{π/3}\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}}dx +\int\limits_{π/6}^{π/3}\frac{\sqrt{cotx}}{\sqrt{cotx}+\sqrt{tanx}}dx

2I = \int\limits_{π/6}^{π/3}(\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}} +\frac{\sqrt{tanx}}{\sqrt{tanx}+\sqrt{cotx}})dx

2I=\int\limits_{π/6}^{π/3}dx

2I=π/6

I=π/12

Question 4. \int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx

Solution:

We know \int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx

So,

\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx = \int\limits_{π/6}^{π/3} \frac{\sqrt{sin(π/2-x)}}{\sqrt{sin(π/2-x)}+\sqrt{cos(π/2-x)}} dx

\int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx = \int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}} dx

if 

I = \int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx

then,

 I = \int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}} dx

2I = \int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}} dx +\int\limits_{π/6}^{π/3}\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}} dx

2I = \int\limits_{π/6}^{π/3}\frac{\sqrt{sinx}+\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}} dx

2I = \int\limits_{π/6}^{π/3}dx

2I=π/6

I=π/12

Question 5. \int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx

Solution:

We know \int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx

so,

\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx =\int\limits_{-π/4}^{π/4} \frac{tan^2{-x}}{1+e^{-x}}dx

\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^{-x}}dx

if 

I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx

then,

 I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^{-x}}dx

2I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx +\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^{-x}}dx

2I=\int\limits_{-π/4}^{π/4} \frac{tan^2x}{1+e^x}dx +\int\limits_{-π/4}^{π/4} \frac{e^xtan^2x}{1+e^x}dx

2I=\int\limits_{-π/4}^{π/4} \frac{tan^2x+e^xtan^2x}{1+e^x}dx

2I=\int\limits_{-π/4}^{π/4} \frac{(e^x+1)tan^2x}{1+e^x}dx

2I=\int\limits_{-π/4}^{π/4} tan^2xdx

I=\frac{\int\limits_{-π/4}^{π/4} tan^2x}{2}dx

we know if 

f(x) is even \int\limits_{-a}^af(x)dx=2\int\limits_{0}^af(x)dx

f(x) is odd \int\limits_{-a}^af(x)dx=0

Here, f(x) = tan2x which is even

hence,

I=\int\limits_{0}^{π/4}tan^2xdx

I=\int\limits_{0}^{π/4}(sec^2x-1)dx

I=\left.({tanx-x})\right|_0^{π/4}

I = I=π/4

Question 6. \int\limits_{-a}^a \frac{1}{1+a^x}dx,a>0

Solution:

We know \int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx

So, 

\int\limits_{-a}^a \frac{1}{1+a^x}dx = \int\limits_{-a}^a \frac{1}{1+a^{-x}}dx

if, I = \int\limits_{-a}^a \frac{1}{1+a^x}dx then I=\int\limits_{-a}^a \frac{1}{1+a^{-x}}dx

So, 

2I=\int\limits_{-a}^a \frac{1}{1+a^x}dx+ \int\limits_{-a}^a \frac{1}{1+a^{-x}}dx

2I=\int\limits_{-a}^a \frac{1}{1+a^x}+ \frac{1}{1+a^{-x}}dx

2I= \int\limits_{-a}^a \frac{1+a^x}{1+a^x}dx

2I= \int\limits_{-a}^a \frac{1}{1+a^x}+ \frac{a^x}{1+a^x}dx

2I= \int\limits_{-a}^a dx

2I=2a

I=a

Question 7. \int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx

Solution:

We know \int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx

Hence, 

\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx =\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{-tanx}} dx

if, I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx

then I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{-tanx}} dx

so,

2I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} dx+\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{-tanx}} dx

2I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} + \frac{1}{1+e^{-tanx}} dx

2I=\int\limits_{-π/3}^{π/3} \frac{1}{1+e^{tanx}} + \frac{e^{tanx}}{1+e^{tanx}} dx

2I=\int\limits_{-π/3}^{π/3} dx

2I=2π/3

I=π/3

Question 8. \int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx

Solution:

We know \int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx

hence, \int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx =\int\limits_{-π/2}^{π/2} \frac{cos^2(-x)}{1+e^{-x}}dx

\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx = \int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^{-x}}dx

if I=\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx

Then, I=\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^{-x}}dx

So, 2I=\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}dx +\int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^{-x}}dx

2I= \int\limits_{-π/2}^{π/2} \frac{cos^2x}{1+e^x}+ \frac{e^xcos^2x}{1+e^x}dx

2I=\int\limits_{-π/2}^{π/2} \frac{cos^2x+e^xcos^2x}{1+e^x}dx

2I=\int\limits_{-π/2}^{π/2} \frac{(1+e^x)cos^2x}{1+e^x}dx

2I=\int\limits_{-π/2}^{π/2} cos^2x dx

2I=\int\limits_{-π/2}^{π/2} \frac{1+cos2x}{2}dx

I= \left.\frac{1}{4}(x+\frac{sin2x}2)\right|_{-π/2}^{π/2}

I=\frac{1}{4}[\frac{π}2-(-\frac{π}2)]

I=\frac{π}4

Question 9.  \int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5+1}{cos^2x} dx

Solution:

\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5+1}{cos^2x} dx

\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5}{cos^2x}+\frac{1}{cos^2x} dx

\int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5}{cos^2x}+\int\limits_{-π/4}^{π/4}sec^2x dx

if f(x) is even 

\int\limits_{-a}^af(x)dx=2\int\limits_{0}^af(x)dx

if f(x) is odd 

\int\limits_{-a}^af(x)dx=0

here,  \int\limits_{-π/4}^{π/4} \frac{x^{11}-3x^9+5x^7-x^5}{cos^2x}dx       is odd and \int\limits_{-π/4}^{π/4}sec^2x dx       

is even

Hence,

0+2\int\limits_0^{π/4}sec^2x dx

\left.2tanx\right|_0^{\frac{π}{4}}

2

Question 10.  \int\limits_{a}^{b} \frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx,n\in N,n\geq 2

Solution:

if 

I=\int\limits_{a}^{b} \frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx

then, 

I=\int\limits_{a}^{b} \frac{{(a+b-x)}^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx

2I=\int\limits_{a}^{b} \frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx +\int\limits_{a}^{b} \frac{(a+b-x)^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx

2I=\int\limits_{a}^{b} \frac{x^{1/n}+(a+b-x)^{1/n}}{x^{1/n}+(a+b-x)^{1/n}} dx

2I=\int\limits_{a}^{b} dx

I=\frac{b-a}{2}

Question 11.  \int\limits_0^{π/2} (2logcosx-logsinx2x)dx

Solution:

let, I=\int\limits_0^{π/2} (2logcosx-logsinx2x)dx

I=\int\limits_0^{π/2} (log\frac{cos^2x}{sinx2x})dx

I=\int\limits_0^{π/2} (log\frac{cos^2x}{2sinxcosx})dx

I=\int\limits_0^{π/2} (log\frac{cosx}{2sinx})dx

I=\int\limits_0^{π/2} logcosxdx-\int\limits_0^{π/2} logsinxdx-\int\limits_0^{π/2}log2dx

we know that, \int\limits_0^{π/2}logcosxdx=\int\limits_0^{π/2}logsinxdx

hence,

I=-\int\limits_0^{π/2}log2dx =\frac{-π}{2}log2

Question 12.  \int\limits_0^a \frac{\sqrt x}{\sqrt x+\sqrt {a-x}} dx

Solution:

Let, I=\int\limits_0^a \frac{\sqrt x}{\sqrt x+\sqrt {a-x}} dx

we know that ,\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx

so, I=\int\limits_0^a \frac{\sqrt {a-x}}{\sqrt{a-x}+\sqrt {x}} dx

then, 

2I= \int\limits_0^a \frac{\sqrt x}{\sqrt x+\sqrt {a-x}} dx + \int\limits_0^a \frac{\sqrt{ a-x}}{\sqrt {a-x}+\sqrt x} dx

2I= \int\limits_0^a \frac{\sqrt x+\sqrt{a-x}}{\sqrt x+\sqrt {a-x}} dx

2I= \int\limits_0^a  dx

2I=a

I=\frac{a}{2}

Question 13.  \int\limits_0^5 \frac{\sqrt[4]{x+4} }{\sqrt[4]{x+4} +\sqrt[4]{9-x} }dx  

Solution:

We know that, \int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx

So,

I=\int\limits_0^5 \frac {\sqrt[4]{(5-x)+4} }{\sqrt[4]{(5-x)+4} +\sqrt[4]{9-(5-x)} }dx  

I=\int\limits_0^5 \frac{\sqrt[4]{9-x} }{\sqrt[4]{9-x} +\sqrt[4]{4+x} }dx  

then,

2I=\int\limits_0^5 \frac{\sqrt[4]{x+4} }{\sqrt[4]{x+4} +\sqrt[4]{9-x} }dx   +\int\limits_0^5 \frac{\sqrt[4]{9-x} }{\sqrt[4]{9-x} +\sqrt[4]{4+x} }dx  

2I= \int\limits_0^5 \frac{\sqrt[4]{x+4}\sqrt[4]{9-x} }{\sqrt[4]{x+4} +\sqrt[4]{9-x} }dx  

2I= \int\limits_0^5 dx  

II=\frac{5}{2}

Question 14.  \int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{7-x}}dx

Solution:

We know that, \int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx

so,

I=\int\limits_0^7 \frac{\sqrt[3]{7-x}}{\sqrt[3]{7-x}+\sqrt[3]{x}}dx

then,

2I=\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{7-x}}dx+ \int\limits_0^7 \frac{\sqrt[3]{7-x}}{\sqrt[3]{7-x}+\sqrt[3]{x}}dx

2I=\int\limits_0^7 dx

I=\frac{7}{2}

Question 15.  \int\limits_{π/6}^{π/3} \frac{1}{1+\sqrt {tanx}}dx

Solution:

We know that, \int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx

Let, I=\int\limits_{π/6}^{π/3} \frac{1}{1+\sqrt {tanx}}dx

I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt {sinx}}dx

hence,

I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cos(\frac{π}{2}- x)}}{\sqrt{cos(\frac{π}{2}-x)} +\sqrt {sin(\frac{π}{2}-x)}}dx

I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt {sinx}}dx

2I=\int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt {sinx}}dx + \int\limits_{π/6}^{π/3} \frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt {sinx}}dx

2I= \int\limits_{π/6}^{π/3} \frac{\sqrt{cosx}+\sqrt{sinx}}{\sqrt{cosx}+\sqrt {sinx}}dx

2I= \int\limits_{π/6}^{π/3} dx

I=\frac{π}{12}

Question 16. If f(a+b-x)=f(x), then prove that \int\limits_a^b xf(x)dx=\frac{a+b}{2} \int\limits_a^bf(x)dx

Solution:

I=\int\limits_a^bf(x)dx

I=\int\limits_a^b(a+b-x)f(a+b-x)dx

I=\int\limits_a^b(a+b-x)f(x)dx........... [\because f(a+b-x)=f(x)]

I=\int\limits_a^b(a+b)f(x)dx-\int\limits_a^bf(x)dx

I=(a+b)\int\limits_a^bf(x)dx-I

2I=(a+b)\int\limits_a^bf(x)dx

I=\frac{(a+b)}{2}\int\limits_a^bf(x)dx

\therefore \int\limits_a^bxf(x)dx=\frac{(a+b)}{2}\int\limits_a^bf(x)dx

Summary

This exercise typically focuses on evaluating definite integrals using various methods and properties. Key concepts often covered include:

  • Basic definite integral evaluation
  • Properties of definite integrals
  • Integration by substitution
  • Integration by parts
  • Evaluating integrals with trigonometric functions
  • Dealing with even and odd functions in definite integrals
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