Evaluating the integral ∫0π41−16sin2xdx I tried a substitution u=4sin⁡x, dx=du16−u2 hence the integration...

minikim38

minikim38

Answered

2022-01-26

Evaluating the integral 0π4116sin2xdx
I tried a substitution
u=4sinx,   dx=du16u2
hence the integration will be
u=0u=221u216u2du
But I could not complete the solution using this substitution.

Answer & Explanation

Dominique Green

Dominique Green

Expert

2022-01-27Added 11 answers

0π4116sin2(x)dx=0π4142sin2(x)dx
=E(π4,4)
E is a Legendre Integral.

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