x 1+ x d x - Vidyadip

Question

$\int \frac{\sin x}{1+\sin x} d x$

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Answer

$\text { Let } I =\int \frac{\sin x}{1+\sin x} d x$
$=\int \frac{\sin x}{1+\sin x} \cdot \frac{1-\sin x}{1-\sin x} d x$
$=\int \frac{\sin x-\sin ^2 x}{1-\sin ^2 x} d x$
$=\int \frac{\sin x-\sin ^2 x}{\cos 2} d x$
$=\int\left(\frac{\sin x}{\cos 2}-\frac{\sin ^2 x}{\cos ^2 x}\right) d x$
$\left.=\int \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x}-\tan ^2 x\right) d x$
$=\int\left[\sec x \tan x d x-\int \sec ^2 x d x+\int 1 \cdot d x\right.$

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