12
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Tunk-Fey May 7 '14 at 5:36
14
$\begingroup$

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. $\qquad\square$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Bennett Gardiner May 7 '14 at 6:37
  • 6
    $\begingroup$ Sorry. If I knew, I would have waited. $\endgroup$ – Omran Kouba May 7 '14 at 6:40
  • $\begingroup$ 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. $\endgroup$ – Nicholas Pipitone May 7 '14 at 7:03
  • $\begingroup$ and the $t$ in the log!. $\endgroup$ – Omran Kouba May 7 '14 at 7:04
5
$\begingroup$

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 $$ \frac{1}{2} \int_0^1 \log (1-w^2)^{1/4} \ \mathrm{d}w = {1\over 8} \int_0^1 \log(1+w) +\log (1-w) \ \mathrm{d}w = \frac{\ln 2 -1}{4}. $$

$\endgroup$
5
$\begingroup$

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

$\endgroup$

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.