16
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – George V. Williams Jan 11 '13 at 23:52
8
$\begingroup$

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 $

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

$\endgroup$
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.