logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

By using the properties of definite integrals,evaluate the integral\[\int\limits_0^\pi\frac{x\;dx}{1+\sin x}\]

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • (i)$\int\limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)$\int \limits_0^af(x)dx=\int \limits_0^af(a-x)dx$
Given $I=\int\limits_0^\pi\frac{x\;dx}{1+\sin x}-----(1)$
 
Applying the property $\int \limits_0^af(x)dx=\int \limits_0^af(a-x)dx$
 
$I=\int\limits_0^\pi\frac{(\pi-x)}{1+\sin(\pi-x)}dx-----(1)$
 
But $\sin(\pi -x)=\sin x$
 
$I=\int\limits_0^\pi\frac{(\pi-x)}{1+\sin x}dx-----(2)$
 
Adding equ (1) and equ(2)
 
$2I=\int\limits_0^\pi\frac{xdx}{1+\sin x}+\int\limits_0^\pi\frac{(\pi-x)}{1+\sin x}dx$
 
On simplifying we get,
 
$2I=\int\limits_0^\pi\frac{\pi}{1+\sin x}dx$
 
Multiply and divide by $(1-\sin x)$
 
$2I=\pi\int\limits_0^\pi\frac{(1-\sin x)}{1+\sin x)(1-\sin x)}dx$
 
$(1+\sin x)(1-\sin x)=1-\sin^2x=\cos ^2x$
 
Therefore $2I=\pi\int\limits_0^\pi\frac{1-\sin x}{\cos^2 x}$
 
On seperating the terms,
 
$2I=\pi\int\limits_0^\pi\large\frac{1}{\cos^2 x}-\int\limits_0^\pi\frac{\sin x}{\cos x}.\frac{1}{\cos x}dx$
 
$=\pi\int\limits_0^\pi \sec ^2x-\int \limits_0^\pi \sec x.\tan x$
 
On integrating we get
 
$2I=\pi\bigg\{[\tan x]_0^\pi -[\sec x]_0^\pi \bigg\}$
 
On applying limits,
 
$2I=\pi[(\tan \pi-\tan 0)-(\sec \pi-\sec 0)]$
 
But $\tan \pi=\tan 0=0$
 
and $\sec \pi=-1\;and\; \sec 0=1$
 
Now substituting these values,
 
$2I=\pi[0-(-1-1)]$
 
$2I=2\pi$
 
Therefore $I=\pi$

 

 

answered Feb 14, 2013 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...