Question

(sinx+cosx)/(sin⁴x+cos⁴x) Integrate with respect to x......

Answer 1

(sinx+cosx) / (sin^2 x+cos^2 x)^2-2sin^2x cos^2x 

=(sinx+cosx) / 1-2sin^2x cos^2x 

=2(sinx+cosx) / 2-(2sinx cosx)^2 

=2(sinx+cosx) / 2-( 1-(sinx-cosx)^2 )^2 

now sub sinx-cosx = u..then cosx+sinx dx =du..numerator 

hence integral is 

I=2du / 2-( 1-(u)^2 )^2 

I=2du / 1+2u^2-u^4.