Question

integrate 1/(sin²x+sin2x)​

Answer 1

Let  I  be the integral,

I=∫1sin(x)+sin(2x)dx  

Using the double-angle identity  sin(2x)=2sin(x)cos(x) ,

I=∫1sin(x)(1+2cos(x))dx  

Now, time for a u-substitution.

Let  u=1+2cos(x)  

u−12=cos(x)  

du=−2sin(x)dx  

∴I=−12∫1sin(x)udusin(x)  

=−12∫1sin2(x)udu  

Using the double-angle identity  sin2(x)=1−cos2(x)  and the relation  u−12=cos(x)  

I=−12∫1u−ucos2(x)du  

=−12∫1u−u(u−12)2du  

=−42∫1u(4−(u−1)2)du  

=−2∫1u(4−u2−1+2u))du  

=−2∫1u(3−u2+2u))du  

Factoring the quadratic in the denominator,

I=−2∫1u(u+1)(−u+3)du