Question
Integrate the function $\frac{x}{1+\sin x}$ with respect to $x$.

Answer

Let $\quad I =\frac{x}{1+\sin x} d x$
On multiplying by $(1-\sin x)$ in numerator and denominator,
$
\begin{aligned}
I & =\int \frac{x(1-\sin x)}{(1+\sin x)(1-\sin x)} d x \\
& =\int \frac{x(1-\sin x)}{1-\sin ^2 x} d x=\int \frac{x(1-\sin x)}{\cos ^2 x} d x \\
& =\int_{\text {I }}^x \sec _{\text {II }}{ }^2 x d x-\int_{I} x \sec x \tan x d x \quad \text { I (using ILATE) }\\ & =x \cdot \tan x-\int 1 \cdot \tan x d x-\quad\left(x \cdot \sec x-\int 1 \cdot \sec x d x\right) \\
& =x \tan x-\int \tan x d x-x \sec x+\int \sec x d x \\
& =x \tan x-\log \sec x-x \sec x \\
& +\log (\sec x+\tan x)+C \\
& =x(\tan x-\sec x)+\log \left(\frac{\sec x+\tan x}{\sec x}\right)+C
\end{aligned}
$

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