Question

Int_Sin4x/Sin^4x+Cos^4x=?

Answer 1

`I=-ln(cos(4x)+3)+C`

Explanation:

We want to solve

`I=intsin(4x)/(sin^4(x)+cos^4(x))dx`

Rewrite the integrand using the identity

`color(blue)(sin^4(x)+cos^4(x)=1/4(cos(4x)+3))`

Thus

`I=intsin(4x)/(1/4(cos(4x)+3))dx=int(4sin(4x))/(cos(4x)+3)dx`

Make a substitution `color(brown)(u=cos(4x)+3=>du=-4sin(4x)dx`

`I=-int1/udu=-ln(u)+C`

Substitute back `u=cos(4x)+3`

`I=-ln(cos(4x)+3)+C`