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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 $$\frac{\pi}{4\sqrt{2}}$$

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    $\begingroup$ 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, 2016 at 12:42

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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.

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  • $\begingroup$ 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, 2016 at 12:59
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    $\begingroup$ @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, 2016 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, 2016 at 7:43

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