Question

integration of secx/(1+cosecx)

Answer 1

Hey there!

$\frac{\sec x}{1+\csc x}=\frac{\tan x}{1+\sin x}=\frac{\tan x(1-\sin x)}{1-\sin^2x}=\frac{1}{\cos^2x}\tan x-\frac{1}{\cos x}\tan^2 x$

Now,

$\int\frac{1}{\cos^2x}\tan x\,dx=\frac{1}{2}\tan^2x$

For the other part we integrate by parts,

[tex]\begin{aligned} \int\frac{1}{\cos x}\tan^2x\,dx&=\int\frac{\sin x}{\cos^2 x}\tan x\,dx\\ &=\frac{1}{\cos x}\tan x-\int\frac{1}{\cos x}(1+\tan^2x)\,dx\\ &=\frac{1}{\cos x}\tan x-\int\frac{\cos x}{1-\sin^2x}\,dx-\int\frac{1}{\cos x}\tan^2x\,dx\\ &=\frac{1}{\cos x}\tan x-\text{artanh}\,\sin x-\int\frac{1}{\cos x}\tan^2x\,dx. \end{aligned}[/tex]

Finally,

$\int\frac{\sec x}{1+\csc x}\,dx=\frac{1}{2}\text{artanh}\,(\sin x)-\frac{1}{2}\frac{1}{\cos x}\tan x+\frac{1}{2}\tan^2x+C$

Cheers!

Answer 2

Answer:

your answer attached in the photo