2
$\begingroup$

I'm trying to evaluate $$\int_{0}^{\frac{\pi}{2}} \sqrt{\sin ^3 x \cos x} dx$$

This looks like an opportunity to isolate a $\sin x$ or $\cos x$ in order to use the reverse chain rule. However, the best I was able to do, in order to achieve this, was reduce the integral to $$\int_{0}^{\frac{\pi}{2}}\sqrt{\sqrt{1- \cos ^2 x} \cos x}\sin xdx$$

which then becomes, under the substitution $u=\cos x$

$$\int_{0}^{1} \sqrt[4]{1-u^2}\sqrt{u}du$$

But I am not quite sure what to do from here. I am not even sure if the integral can be found via elementary means, though Wolfram gives a closed form of $$\frac{\pi}{4\sqrt{2}}$$

$\endgroup$
  • $\begingroup$ If you know the beta function, the integral is $\dfrac 1 2 B\left (\dfrac 5 4, \dfrac 3 4 \right)$. $\endgroup$ – M. Vinay Apr 13 '16 at 12:42
2
$\begingroup$

If you do not want to use the Beta function, you could set $$ u=\sqrt{\tan x}. $$ That will transform your integral into $$ 2\int_0^{+\infty}\frac{u^4}{(1+u^4)^2}\,du $$ which can be handled via partial fraction decomposition or with complex methods. The former is rather tedious. I leave those calculations for you.

$\endgroup$
  • $\begingroup$ I am able to do the rest, probably contour integration would be easiest. However, I'm curious as to how you knew to make that particular substitution. It seems rather unnatural. $\endgroup$ – Trogdor Apr 13 '16 at 12:59
  • 2
    $\begingroup$ @Trogdor Perhaps not so unnatural if you first try different rearrangements of the integrand, in particular, rewriting $\sin^3 x \cos x$ (which is somewhat ugly because of the sine-cosine imbalance), as $\dfrac{\tan^3 x}{\sec^2 x} = \dfrac{\tan^3 x}{1 + \tan^2 x}$. $\endgroup$ – M. Vinay Apr 13 '16 at 13:02
  • $\begingroup$ Indeed. In my calculations I first did $s=\tan x$, but then it appeared to be good to put $u=\sqrt{s}$. In the answer the substitutions got combined. (fixed a misprint) $\endgroup$ – mickep Apr 14 '16 at 7:43

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.