How to integrate $\csc (2x)$ by using substitution $t=\tan x$ ? I know a different way to solve it. But the book mentioned must be using substitution of tan x
$\endgroup$ 11 Answer
$\begingroup$$cosec(2x) = \frac{1}{sin(2x)} = \frac{1}{2sin(x)cos(x)} = \frac{cosx}{2sin(x)cos^{2}x} = \frac{sec^{2}x}{2\frac{sinx}{cosx}} = \frac{1+tan^2x}{2tanx}$
$$\int cosec(2x) dx = \int \frac{1+tan^2x}{2tan(x)} dx$$
Now put $t = tan(x) \implies dt = (sec^2x)dx = (1+tan^2x)dx = (1+t^2)dx$
Hence the integral becomes
$$\int \frac{1+t^2}{2t} \times \frac{dt}{1+t^2}$$
Can you complete the solution?
$\endgroup$ 5
