Question

Integration help?

Answer 1

\[\LARGE \int~\frac{dx}{sinx+tanx}\]

Answer 2

\[sinx =\frac{ 2\tan \frac{ x }{ 2 } }{ 1+{\tan^2 \frac{ x }{ 2 }} }\\tan x=\frac{ 2\tan \frac{ x }{ 2 } }{ 1-{\tan^2 \frac{ x }{ 2 }} }\]

Answer 3

if you put them and simplify
then use u=tan x/2
you will have (1-u^2)/2u in integral
and it is easy to solve

Answer 4

That's actually what I did, and got the same thing, I thought I did something wrong tho. I separate and integrate right?

Answer 5

\[\int\limits_{?}^{?}\frac{ 1-u^2 }{ 2u }du=0.5\int\limits_{?}^{?}(\frac{ 1 }{ u }-u)du\]

Answer 6

Thank you :)