Question

integrate : 1/sin^4x+cos^4x dx

Answer 1

1/(cos^4x +sin^4x)

Dividing numerator and denominator by cos^4x

= sec^4x/(1+tan^4x )

Taking substitution t=tanx,sec^4xdx =dt and 1+tan^2x =sec^2x

=(1+t^2).dt/(1+t^4)

Now dividing numerator and denominator by t^2

= (1/t^2 + 1).dt/(1/t^2 +t^2)

Now (1/t^2 +t^2 ) =(t-1/t)^2 +2,taking substitution

t-1/t =m ,1+1/t^2 = dm

= dm/(m^2+2)

=1/(2^1/2)×(arctan(m/(2^1/2) ) +C

replacing m by t-1/t and t by Tanx we get the expression below

1/(2^1/2) ×(tan((tanx-cotx)/2^1/2 ) +C