Evaluate: `int1/(sin^4x+cos^4x)dx`

Answer

`(1)/(sqrt(2))"tan"^(-1)((tan^(2)x-1)/(sqrt(2)tanx))+C`

Explanation

`I=int(1)/(sin^(4)x+cos^(4)x)dx=int(1//cos^(4)x)/((sin^(4)x+cos^(4)x)/(cos^(4)x))dx`
`=int(sec^(4)x)/(tan^(4)x+1)dx=int(sec^(2)x sec^(2)x)/(tan^(4)x+1)dx`
`=int((1+tan^(2)x)/(1+tan^(4)x))sec^(2)x dx`
` "Putting " tan x=t " and "sec^(2)x dx=dt, " we get" `
`I=int(1+t^(2))/(1+t^(2))dt=(1)/(sqrt(2))"tan"^(-1)((t^(2)-1)/(sqrt(2)t))+C`
`=(1)/(sqrt(2))"tan"^(-1)((tan^(2)x-1)/(sqrt(2)tanx))+C`