Question

Evaluate:
$\int \frac{dx}{\sin x+\cos x}$

Answer 1

$\int \frac{dx}{\sin x+\cos x}$

$\sin x.\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}};\cos x,\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$

$\therefore \int \frac{dx}{\sin x+\cos x}$

$=\int \frac{1+{\tan }^2\frac{x}{2}}{2\tan \frac{x}{2}+1-{\tan }^2\frac{x}{2}}.dx$

Put $\tan \frac{x}{2}=u\Rightarrow dx=\frac{2}{1+u^2}du$

$\Rightarrow \int \frac{dx}{\sin x+\cos x}=\int \frac{1+u^2}{2u+1-u^2}.\frac{2}{1+u^2}du$

$=\int \frac{2}{2u+1-u^2}du$

$=-2\int \frac{1}{(u-\sqrt{2}-1)(u+\sqrt{2}-1)}du$

$=\frac{-2.1}{2^3/2}\int \frac{1}{u-\sqrt{2}-1}du-\int \frac{1}{u+\sqrt{2}-1}du$

$=\frac{ln(u+\sqrt{2}-1)}{\sqrt{2}}-\frac{ln(4-\sqrt{2}-1)}{\sqrt{2}}$

$=\frac{ln\left(|\tan \frac{x}{2}+\sqrt{2}-1|\right)-ln\left(|\tan \frac{x}{2}-\sqrt{2}+1|\right)}{\sqrt{2}}+c$

Hence, solved.