I want to calculate $$ \int_0^{\pi/2}\tan(x)\ln(\sin(x))\,dx. $$

I tried taking $\cos x$ as $t$ and converting the whole expression in $t$ and then integrating by parts. That didn't help. Integrating by parts directly also doesn't help. I'm guessing there must be a different, better approach to solve this.

Can someone help?

  • $\begingroup$ Never mind... This is the derivative of the beta function in disguise. The final result is $-\dfrac{\pi^2}{24}$ $\endgroup$ – Lucian Oct 11 '15 at 6:56
  • $\begingroup$ Confusing, could you please update your question with the actual integral you want to calculate? $\endgroup$ – mickep Oct 11 '15 at 7:05
  • $\begingroup$ I didn't quite get that beta function answer. Is there any way to do it besides that? This is a question given in the book provided by my coaching institute where we haven't yet been taught Beta functions. So there must be another way of doing it. Although yeah, the answer is correct. As for the king property, yes. That doesn't help either. $\endgroup$ – Ashish Gupta Oct 11 '15 at 7:07
  • $\begingroup$ @mickep, this is the actual integral. I didn't get here using some substitution. As in, this IS the question given. $\endgroup$ – Ashish Gupta Oct 11 '15 at 7:08
  • $\begingroup$ If instead of $\tan(x)$, $\cot(x)$ would have been there, the integral could have been solved within seconds. $\endgroup$ – Aditya Agarwal Oct 11 '15 at 7:44

\begin{align} I&=\int^{\frac\pi2}_0\tan(x)\ln(\sin(x))dx \\ &=\int_{0}^1\frac{\ln(\sqrt{1-t^2})}tdt\tag{$t=\cos{x}$} \\ &=\int_{0}^1\frac{y\ln y}{1-y^2}\:dy\tag{$y=\sqrt{1-t^2}$} \\ &=\int_{0}^1y\ln y\left(\sum_{n=0}^{\infty}y^{2n}\right)dy \\ &=\sum_{n=0}^{\infty}\int_{0}^1y^{2n+1}\ln y\:dy\tag1 \\ &=-\sum_{n=0}^{\infty}\frac1{(2n+2)^2} \\ &=-\frac{\pi^2}{24} \end{align} Note in $(1)$ $$ \int_{0}^1y^{2n+1}\ln y\:dy=\frac{y^{2n+2}\ln y}{2n+2}\bigg |_0^1-\frac1{2n+2}\int_{0}^1y^{2n+1}\:dy=-\frac1{(2n+2)^2} $$

  • $\begingroup$ How did you introduce summation? $\endgroup$ – Aditya Agarwal Oct 11 '15 at 7:42
  • $\begingroup$ By taylor series of $\frac1{1-y^2}$ $\endgroup$ – Math Wizard Oct 11 '15 at 7:43
  • $\begingroup$ I do, and this infact is a better solution for school purposes. If nothing else, I'll accept this answer. I'll wait for a different method for a while though. $\endgroup$ – Ashish Gupta Oct 11 '15 at 7:52

Hint: $$I=\int^{\frac\pi2}_0\tan(x)\ln(\sin(x))dx$$ Put $\cos x=t; -\sin xdx=dt$
$$=-\int^{0}_1\frac{\ln(\sqrt{1-t^2})}tdt$$ $$=\frac12\int^1_0\frac{2t\ln(\sqrt{1-t^2})}{t^2}dt$$ Put $1-t^2=u;-2tdt=du$ $$\frac12\int^1_0\frac{\ln(\sqrt u)}{u-1}du$$ $$\frac12\int^1_0\frac{\ln(\sqrt u)}{u-1}du=\frac12[-\frac{\text{Li}_2(1-u)}{2}]^1_0=-\frac{\pi^2}{24}$$

  • $\begingroup$ Doesn't help. In the second integral, you again get integration of log.. $\endgroup$ – Ashish Gupta Oct 11 '15 at 7:17
  • $\begingroup$ One minute, updating. $\endgroup$ – Aditya Agarwal Oct 11 '15 at 7:17
  • $\begingroup$ I think there is a typo in your question. $\endgroup$ – Aditya Agarwal Oct 11 '15 at 7:19
  • $\begingroup$ No, there isn't. Lucian got the correct answer to this question, albeit using a method that we can't use/hasn't been taught to us.. $\endgroup$ – Ashish Gupta Oct 11 '15 at 7:20
  • $\begingroup$ See my edit then. I said that there was a typo because I don't think you can simplify it using elementary functions. $\endgroup$ – Aditya Agarwal Oct 11 '15 at 7:22

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