Question

$$\int \frac{1}{1-\sin x+\cos x}\text{ dx }$$

Answer 1

$$\text{ Let I }=\int \frac{1}{1-\sin x+\cos x}dx$$
$$\text{ Putting sin x}=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\text{ and }cosx=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$$
$$=\int \frac{1}{1-\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}+\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}dx$$
$$=\int \frac{(1+{\tan }^2\frac{x}{2})}{(1+{\tan }^2\frac{x}{2})-2\tan x(2+1-{\tan }^2\frac{x}{2})}dx$$
$$=\int \frac{{\sec }^2\frac{x}{2}}{2-2\tan \left(\frac{x}{2}\right)}dx$$
$$=\frac{1}{2}\int \frac{{\sec }^2\left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}dx$$
$$Let[1-\tan \left(\frac{x}{2}\right)]=t$$
$$\Rightarrow -{\text{ sec}}^2\left(\frac{x}{2}\right)\times \frac{1}{2}dx=dt$$
$$\Rightarrow {\text{ sec}}^2\left(\frac{x}{2}\right)dx=-\text{ 2dt }$$
$$\therefore I=\frac{1}{2}\int \frac{-2dt}{t}$$
$$=-\int \frac{dt}{t}$$
$$=-\text{ ln }\left|t\right|+C$$
$$=-\text{ ln }|1-\tan \frac{x}{2}|+C$$