Question

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

Answer 1

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

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

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

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

Let $\tan \frac{x}{2}=t$ $\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$

$\Rightarrow {\sec }^2\frac{x}{2}dx=2dt$

Replacing the values, we get

$I=\int \frac{dt}{1-t^2+2t}$

$=\int \frac{2dt}{1-(t^2-2t)}$

$=\int \frac{2dt}{2-(t-1{)}^2}$

$=2\int \frac{dt}{(\sqrt{2}{)}^2-(t-1{)}^2}$

$=\frac{1}{\sqrt{2}}\log \left|\frac{\sqrt{2}+(t-1)}{\sqrt{2}-(t-1)}\right|+C$

$=\frac{1}{\sqrt{2}}\log \left|\frac{\sqrt{2}+\tan \frac{x}{2}-1}{\sqrt{2}-\tan \frac{x}{2}+1)}\right|+C$