Evaluate: x 1+ x d x - Vidyadip

Question

Evaluate:
$\int \frac{\sin x}{1+\sin x} \cdot d x$

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Answer

$ \int \frac{\sin x}{1+\sin x} d x$
$= \int \frac{\sin x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x} d x$
$= \int \frac{\sin x-\sin ^2 x}{1-\sin ^2 x} d x$
$= \int \frac{\sin x-\sin ^2 x}{\cos ^2 x} d x$
$=\int\left(\frac{\sin x}{\cos ^2 x}-\frac{\sin ^2 x}{\cos ^2 x}\right) d x$
$=\int\left(\frac{1}{\cos x}\right)\left(\frac{\sin x}{\cos x}\right) d x-\int \tan ^2 x d x$
$=\int \sec x \tan x d x-\int\left(\sec ^2 x-1\right) d x$
$=\int \sec x \tan x d x-\int \sec ^2 x d x+\int 1 d x$
$=\sec x-\tan x+x+c .$

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