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