Evaluate : $$I = \int_{0}^{\Large\frac\pi2} (\sqrt{\sin x}+\sqrt{\cos x})^{-4}\ dx$$

Attempt : \begin{align} I&=\int_{0}^{\Large\frac\pi2} (\sqrt{\sin x}+\sqrt{\cos x})^{-4}\ dx\\ &=\int_{0}^{\Large\frac\pi2} \left( \frac{\sqrt{\sin x}-\sqrt{\cos x}}{\sin x - \cos x}\right)^{4}\ dx \end{align}

Is it the right way to proceed it? Please guide me. Thanks.


1 Answer 1


\begin{align} \int_0^{\Large\frac\pi2} (\sqrt{\sin x}+\sqrt{\cos x})^{-4}\ dx&=\int_0^{\Large\frac\pi2}\frac1{(\sqrt{\cos x})^4\ (1+\sqrt{\tan x})^4}\ dx\\ &=\int_0^{\Large\frac\pi2}\frac{\sec^2x}{(1+\sqrt{\tan x})^4}\ dx\\ &\stackrel{\color{red}{[1]}}=\int_0^\infty\frac{du}{(1+\sqrt{u})^4}\\ &\stackrel{\color{red}{[2]}}=2\int_0^\infty\frac{t}{(1+t)^4}\ dt\\ &\stackrel{\color{red}{[3]}}=2\cdot\text{B}(2,2)\\ &=\large\color{blue}{\frac13}. \end{align}

Notes :

$\color{red}{[1]}\;\;\;u=\tan x$


$\color{red}{[3]}\;\;\;$Beta function : $\displaystyle\text{B}(x,y)=\int_0^\infty\frac{t^{x-1}}{(1+t)^{x+y}}\ dt$ for $\Re(x)>0$ and $\Re(y)>0$

  • 3
    $\begingroup$ +1 for a very nice answer, but I just wanted to point out for the readers who may be afraid of advanced topics like the Beta function can certainly avoid it by using the fact that $\frac{t}{(1+t)^4}=\frac{1}{(1+t)^3}-\frac{1}{(1+t)^4}$ after the fourth line instead. $\endgroup$
    – David H
    Jul 14, 2014 at 17:40
  • 3
    $\begingroup$ Setting $\tan x=t^2$ will be a shortcut. $\endgroup$
    – Tunk-Fey
    Jul 14, 2014 at 17:44
  • $\begingroup$ do you think this could be generalized to $$\int_0^{\pi/2}\left(\sqrt{\sin x}+\sqrt{\cos x}\right)^{-n}dx?$$ $\endgroup$
    – clathratus
    Mar 20, 2019 at 3:27

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