# Integral $\int_0^{\pi/2} \frac{\sin^3 x\log \sin x}{\sqrt{1+\sin^2 x}}dx=\frac{\ln 2 -1}{4}$

Hi I am trying to prove$$I:=\int_0^{\pi/2} \frac{\sin^3 x\log \sin x}{\sqrt{1+\sin^2 x}}dx=\frac{\ln 2 -1}{4}.$$ Thanks.

I am possibly trying to simplify this to obtain something like $2\int_0^{\pi/2} \log \sin x\, dx=-\pi \ln 2$ since this is easily integrable. However when I try to simplify the terms $$\frac{\sin^3 x}{\sqrt {1+\sin^2 x}}$$ I obtain a more complicated integrand. I am not sure how else to go about this one. I was trying to possibly write $$I(a)=\int_0^{\pi/2} \frac{\sin^3 a x\log \sin x}{\sqrt{1+\sin^2 x}}dx,\quad I'(a)=\int_0^{\pi/2} \frac{\partial}{\partial a}\left(\frac{\sin^3 ax\log \sin x}{\sqrt{1+\sin^2 x}}\right)\, dx,$$ but this didn't simplify anything for me.

I also tried the substitution $y=\sin^2 x$, but couldn't manage to get an integral because of the $\sin 2x$ from the derivative.

• Perhaps you can use IBP by taking $u=\ln\sin x$? Sorry I can't give you a complete answer for this 'claim' because I'm in the middle of the class now. :P – Tunk-Fey May 7 '14 at 5:36

The change of variables $t=\sin x$ yields $$I=\int_0^1\frac{t^3\ln t}{\sqrt{1+t^2}}\frac{dt}{\sqrt{1-t^2}} =\int_0^1\frac{t^3\ln t}{\sqrt{1-t^4}} dt$$ Then setting $t^4=u$ simplifies to \eqalign{ I&=\frac{1}{16}\int_0^1\frac{1}{\sqrt{1-u}}\ln u\, du\cr &=\left[\frac{1}{8}(1-\sqrt{1-u})\ln u\right]_0^1-\frac{1}{8}\int_0^1\frac{1-\sqrt{1-u}}{u} du\cr &=-\frac{1}{8}\int_0^1\frac{1}{1+\sqrt{1-u}} du;\qquad v\leftarrow1+\sqrt{1-u}\cr &=\frac{1}{4}\int_1^2\frac{1-v}{v} du=\frac{\ln 2-1}{4}.} and we are done. $\qquad\square$

• I was just trying to connect to wifi to post my answer and you beat me! I hate how quick you have to be on this website. – Bennett Gardiner May 7 '14 at 6:37
• Sorry. If I knew, I would have waited. – Omran Kouba May 7 '14 at 6:40
• How did the t substitution call for a 1/16 factor? Deriviating $u^4$ is $4u^3$ so shouldn't 1/4 be the balancing factor? -EDIT: Never mind I forgot the ln needed to be stripped of its 1/4 exponent. Silly me. – Nicholas Pipitone May 7 '14 at 7:03
• and the $t$ in the log!. – Omran Kouba May 7 '14 at 7:04

The integral screams for the substitution $u = \sin x$, which transforms it into $$\int_0^1\frac{u^3\log u}{\sqrt{1-u^4}} \ \mathrm{d}u,$$ another, trickier substitution, $w^2 = 1-u^4$ gives $$\frac{1}{2} \int_0^1 \log (1-w^2)^{1/4} \ \mathrm{d}w = {1\over 8} \int_0^1 \log(1+w) +\log (1-w) \ \mathrm{d}w = \frac{\ln 2 -1}{4}.$$

We can derive a more general result:

Consider the integral

\begin{align} I(a)&=\int_0^{\pi/2}\, \frac{\sin(x)^a}{\sqrt{1+\sin(x)^2}}\, dx\\ &=\int_0^{1}\, \frac{t^a}{\sqrt{1-t^4}}\, dt \tag{subst. $t=\sin(x)$}\\ &=\frac{1}{4}\int_0^{1}\, u^{(a-3)/4} (1-u)^{-1/2}\, du \tag{subst. $t^4=u$}\\ &=\frac{1}{4}\mathrm B\left(\frac{a+1}{4},\frac{1}{2}\right) \tag{1} \end{align}

The third line represents a form of Beta function.

\begin{align} \therefore I'(a)&=\int_0^{\pi/2}\, \frac{\sin(x)^a\, \log{\sin{x}}}{\sqrt{1+\sin(x)^2}}\, dx\\ &=\frac{1}{16} \, {\left(\psi_0\left(\frac{a+1}{4} \right)-\psi_0\left(\frac{a+3}{4} \right) \right)} {\rm B}\left(\frac{a+1}{4},\frac{1}{2}\right) \tag{$\frac{d}{da} (1)$}\\ \implies I'(3)&=\frac{\log{2}-1}{4} \end{align}

References: Beta function and Polygamma function