minikim38

2022-01-26

Evaluating the integral ${\int }_{0}^{\frac{\pi }{4}}\sqrt{1-16{\mathrm{sin}}^{2}x}dx$
I tried a substitution

hence the integration will be
${\int }_{u=0}^{u=2\sqrt{2}}\frac{\sqrt{1-{u}^{2}}}{\sqrt{16-{u}^{2}}}du$
But I could not complete the solution using this substitution.

Dominique Green

Expert

${\int }_{0}^{\frac{\pi }{4}}\sqrt{1-16{\mathrm{sin}}^{2}\left(x\right)}dx={\int }_{0}^{\frac{\pi }{4}}\sqrt{1-{4}^{2}{\mathrm{sin}}^{2}\left(x\right)}dx$
$=E\left(\frac{\pi }{4},4\right)$
E is a Legendre Integral.

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