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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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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