Question

Evaluate : `int_0^(pi//4) (sin2x)/(sin^4x + cos^4x)*dx`

Answer 1

Let I = `int_0^(pi//4) (sin2x)/(sin^4x + cos^4x)*dx`

= `int_0^(pi//4) (2sinx cosx)/(sin^4x + cos^4x)*dx`
Dividing each term by cos⁴x, we get

I = `int_0^(pi//4) ((2 sinxcancelcosx)/(cos^4x))/((sin^4x)/(cancelcos^4x )+ 1)*dx`

= `int_0^(pi//4) (2sinx/cosx*1/cos^2)/((tan^2x)^2 + 1)*dx`

 = `int_0^(pi//4) (2tanx*sec^2)/ (tan^4 x+ 1)dx`
Put tan²x = t
∴ 2tanx sec²x·dx = dt
When x = 0, t = tan²0 = 0

When x = `pi/(4), t = tan^2  pi/(4)` = 1

∴ I = `int_0^1 1/(1 + t^2)*dt`

∴ I = `int_0^1 [tan^-1t]_0^1`

= `[tan^-1 t]_0^1`

= tan^–11 – tan^–10

= I =  `pi/(4) - 0`

= I =  `pi/(4)`.