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Home  >>  CBSE XII  >>  Math  >>  Integrals

Find the integrals of the functions\[\frac{\sin^2x}{1+\cos x}\]

$\begin{array}{1 1}x-\sin x+c \\x+\sin x+c \\ x-\cos x+c \\ x+\cos x+c \end{array}$

1 Answer

Toolbox:
  • $(i)\sin2x=2\sin x\cos x.$
  • $\cos x=2\cos^2\frac{x}{2}-1.$
  • $(iii)\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}.$
  • $(iv)\cos x=1-2\sin^2\frac{x}{2}$.
Given:$I=\int\frac{\sin^2x}{1+\cos x}.$
 
This can be written as,
 
$\sin^2x=2\sin^2\frac{x}{2}\cos^2\frac{x}{2}.$
 
$1+\cos x=2\cos^2\frac{x}{2}-1$
 
$I=\int\frac{(2\sin\frac{x}{2}\cos\frac{x}{2})^2}{2\cos^2\frac{x}{2}}dx.$
 
Cancelling $\cos\frac{x}{2}$ which is the common term,we get
 
$\;\;\;=\int2\sin^2\frac{x}{2}dx.$
 
But $\sin^2\frac{x}{2}=\frac{1-\cos x}{2}$.
 
$I=\int 2[\frac{1-\cos x}{2}]dx.$
 
$\;\;\;=\int(1-\cos x)dx.$
 
separating the terms
 
$\;\;\;=\int dx-\int\cos xdx.$
 
On integrating we get,
 
$x-\sin x+c.$
 

 

answered Jan 31, 2013 by sreemathi.v
 
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