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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Find the integrals of the functions\[\frac{\sin^2x}{1+\cos x}\]

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

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  • $(i)\sin2x=2\sin x\cos x.$
  • $\cos x=2\cos^2\frac{x}{2}-1.$
  • $(iii)\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}.$
  • $(iv)\cos x=1-2\sin^2\frac{x}{2}$.
Given:$I=\int\frac{\sin^2x}{1+\cos x}.$
This can be written as,
$1+\cos x=2\cos^2\frac{x}{2}-1$
Cancelling $\cos\frac{x}{2}$ which is the common term,we get
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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