Evaluating the integral ∫0π41−16sin2xdx I tried a substitution u=4sin⁡x, dx=du16−u2 hence the integration will be ∫u=0u=221−u216−u2du But I could not complete the solution using this substitution.

∫0π41−16sin2(x)dx=∫0π41−42sin2(x)dx =E(π4,4) E is a Legendre Integral.

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minikim38

2022-01-26

Evaluating the integral

\int_{0}^{\frac{π}{4}} \sqrt{1 - 16 \sin^{2} x} d x

I tried a substitution

u = 4 \sin x, d x = \frac{d u}{\sqrt{16 - u^{2}}}

hence the integration will be

\int_{u = 0}^{u = 2 \sqrt{2}} \frac{\sqrt{1 - u^{2}}}{\sqrt{16 - u^{2}}} d u

But I could not complete the solution using this substitution.

Dominique Green

Beginner2022-01-27Added 11 answers

\int_{0}^{\frac{π}{4}} \sqrt{1 - 16 \sin^{2} (x)} d x = \int_{0}^{\frac{π}{4}} \sqrt{1 - 4^{2} \sin^{2} (x)} d x

= E (\frac{π}{4}, 4)

E is a Legendre Integral.

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