Question

$\int \frac{2}{1+\tan y}dy$

Answer 1

$\int \frac{2dy}{1+\tan y}$

Let $u=\tan y\Rightarrow du={\sec }^{2}ydy=(1+u^2)dy$

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

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

$=\left[\int \frac{du}{1+u^2}-\frac{1}{2}\int \frac{2du}{1+u^2}+\int \frac{du}{1+u}\right]$

$=[{\tan }^{-1}u-\frac{1}{2}\log (1+u^2)+\log (1+u)]+c$

$={\tan }^{-1}\tan y-\frac{1}{2}\log (1+{\tan }^{2}y)+\log (1+\tan y)+c$ where $u=\tan y$

$=y-\frac{1}{2}\log \left({\sec }^{2}y\right)+\log (1+\tan y)+c$

$=y-\log \sec y+\log (1+\tan y)+c$

$=y+\log \left(\frac{1+\tan y}{\sec y}\right)+c$