wat will be d integration of....

(sinx)/(sin4x)..........???

Dear vats

∫(sinx)/(sin4x) dx

=∫1/4cosxcos2x  dx

=1/4 ∫cosx dx /cos²xcos2x

==1/4 ∫cosx dx /(1-sin²x)(1-2sin²x)

let sinx =t

   cosx dx =dt

=1/4 ∫dt /(1-t²)(1-2t²)

use partial fraction

=-1/4 ∫dt /(1-t²)   + 1/8 ∫dt /(1-2t²)

now you can easly calculate it

Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed  solution very  quickly.

 We are all IITians and here to help you in your IIT JEE preparation.

All the best.
 
Regards,
Askiitians Experts
Badiuddin