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How can we show that $ \displaystyle \int_{0}^{\pi/2}\frac{1}{\sqrt{\sin{x}}}\;{dx}=\int_{0}^{\pi/2}\frac{2}{\sqrt{2-\sin^2{x}}}\;{dx}? $

It feels like it should be simple, but I've tried many things and no luck.

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Note that this is not a general result, $\int \frac{1}{\sqrt{\sin{x}}}\;{dx} \neq \int \frac{2}{\sqrt{2-\sin^2{x}}}\;{dx}$. I assume you want a result that doesn't involve actually evaluating this integral? – George V. Williams Jan 11 '13 at 23:52
up vote 8 down vote accepted

Let $\sin (x) = u $, we get $$ \int_0^1 \frac{1}{\sqrt u \sqrt{1 - u^2}}du = \int_0^1 \frac{1}{\sqrt u \sqrt{(1 - u)(1+u)}}du$$ Again let $u = \cos ^2 y $

$$ \int_{\pi/2}^0 \frac{-2 \cos (y)\; \sin (y) \;dy}{\cos (y) \; \sin (y) \sqrt{1 + \cos^2 y}} = \int_0^{\pi \over 2} \frac{2}{\sqrt{2 - \sin^2 x}} dx $$

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nice prove........ – juantheron Nov 6 '13 at 1:42

To transform the RHS into the LHS: Let $u=\cos^2(x)$, then $u=\sin(t)$.

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