Question

integrate the function [1/sec²x(1 - tanx)²].dx

Answer 1

HELLO DEAR,

it seems there is typing mistake
the correct question is:
integrate the function [sec²x/(1 - tanx)²].dx

now,
given function is $\bold{\int{sec^2x/(1 + tanx)^2}\,dx}$

let (1 - tanx) = t $\bold{\Rightarrow dt/dx = -sec^2x}$

$\bold{\Rightarrow dx = -dt/sec^2x}$

so, $\bold{\Rightarrow \int{sec^x/(1 - tanx)^2}*dt/(-sec^2x)}$

$\bold{\Rightarrow -\int{1/t^2}}$

$\bold{\Rightarrow 1/t + c}$

put the value of t in above function.

$\bold{\Rightarrow 1/(1 + tanx) + c}$

I HOPE ITS HELP YOU,
THANKS