Question #7c0a6

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Answer 1 · 2017-10-30T06:56:26

$pi^2/8$

This integral can be approached by considering a suitable u substitution

Let $u=sin^-1x$
We can differentiate this implicitly;
$sinu = x$
$cosu*(du)/dx = 1$
$(du)/dx = 1/cosu$

Now considering how $sinu = x$ and how $cos^2x+sin^2x=1$
So hence $cosu = root2 (1-x^2)$

Hence $du = (dx)/(root2 ( 1-x^2))$

Hence the intergal becomes $int u du$
$Hence; (1/2)u^2 + c$

Hence the antiderivative is $1/2 (sin^-1x)^2 +c$

Hence evaluating from 0 to 1 we get;

$1/2 (pi/2)^2 = pi^2/8$