This question already has an answer here:

$$\int^{ \pi /2}_{0} \ln (\sin x)\ dx$$

The answer is $- \frac{\pi}{2} \ln 2$.

I have changed it into $$\frac{1}{2} \int^{1}_{0} t d \ln t^{2}$$

But I didn't get the answer with it.


marked as duplicate by Santosh Linkha, user122283, user147263, J. W. Perry, qwr May 31 '14 at 4:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 5
    $\begingroup$ This is a classic! Consider $$I=\int^{\pi/2}_{0}\ln\sin x=\int^{\pi/2}_{0}\ln\cos x$$ Then, ask for $2I$ summing both expressions. Magic will happen. $\endgroup$ – chubakueno May 31 '14 at 2:51
  • $\begingroup$ @chubakueno you should put that in an answer, and possibly the relevant trigonometric identity! $\endgroup$ – DanZimm May 31 '14 at 2:58
  • $\begingroup$ @DanZimm Glacier has already posted it fully worked. I will leave it there as a hint for the ones that don't want to scroll to see the full solution :) $\endgroup$ – chubakueno May 31 '14 at 3:06
  • $\begingroup$ @chubakueno ah good call, didn't realize it at first! $\endgroup$ – DanZimm May 31 '14 at 3:49

Hint: Try

\begin{align}\int^{ \pi /2}_{0} \ln \sin x \, dx &= \int^{ \pi /2}_{0} \ln \left(2 \sin \frac{x}{2} \cos \frac{x}{2} \right) \, dx \\ &=\int^{ \pi /2}_{0} \ln 2 \, dx + \int^{ \pi /2}_{0}\ln \left( \sin \frac{x}{2} \right) \, dx + \int^{ \pi /2}_{0} \ln \left( \cos \frac{x}{2} \right) \, dx \\ &= \frac{\pi}{2} \ln 2 + \underbrace{2 \int^{ \pi /4}_{0} \ln (\sin u) \, du}_{\text{Let }u=x/2} + \underbrace{2 \int^{ \pi /4}_{0} \ln (\cos u) \, du}_{\text{Let }u=x/2} \end{align}


Let $u=\ln\sin x$, $dv=1$. Then: \begin{eqnarray} \underline{u}&\underline{v}\\ \ln\sin x & 1\\ \searrow^+\\ \tan x& \xleftarrow{-} x \end{eqnarray} Thus, $$\int^{\pi/2}_0 \ln\sin xdx=\left.x\ln\sin x\right|^{\pi/2}_0-\int^{\pi/2}_0 x\tan xdx$$ $\int x\tan xdx$ is not expressible by elementary functions, so for your purposes, @Glacier's answer should work.

  • $\begingroup$ After the last edit there's still a type I believe when evaluating $uv$ at the boundary. $\endgroup$ – DanZimm May 31 '14 at 2:52
  • $\begingroup$ @DanZimm Where? $\endgroup$ – user122283 May 31 '14 at 2:52
  • $\begingroup$ The $\ln \sin x$ in the first term should be $x \ln \sin x$ $\endgroup$ – DanZimm May 31 '14 at 2:54

My Solution:: Given $\displaystyle \int_{0}^{\frac{\pi}{2}}\ln (\sin x)dx$

Now Let $\displaystyle \sin x = \left(\frac{e^{ix}-e^{-ix}}{2i}\right) = \frac{1}{2i}\cdot \left(\frac{e^{2ix}-1}{e^{ix}}\right)$

So Integral Convert into $$\displaystyle \int_{0}^{\frac{\pi}{2}}\ln \left(\frac{e^{2ix}-1}{2i\cdot e^{ix}}\right)dx = \int_{0}^{\frac{\pi}{2}}\ln(e^{2ix}-1)dx-\int_{0}^{\frac{\pi}{2}}\ln(e^{ix})dx-\int_{0}^{\frac{\pi}{2}}\ln(2)dx-\int_{0}^{\frac{\pi}{2}}\ln(i)dx$$

$$\displaystyle = \int_{0}^{\frac{\pi}{2}}\ln(e^{2ix}-1)dx-i\cdot \frac{\pi^2}{8}-\frac{\pi}{2}\cdot \ln(2)-i\cdot \frac{\pi^2}{4}$$ Now we will Solve $$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\ln(e^{2ix}-1)dx$$

Using $\displaystyle e^{2ix}-1=\cos(2x)+i\sin(2x)-1 = i\cdot 2\sin x\cdot \cos x-2\sin^2 (x) = 2i\sin x\cdot (\cos x+i\sin x) = 2i\sin x\cdot e^{ix}$

So $$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\ln(2i\cdot \sin x)dx+\int_{0}^{\frac{\pi}{2}}\ln(e^{ix})dx$$

So $$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\ln(2i\cdot \sin x)dx+i\cdot \frac{\pi^2}{8}$$

So $$\displaystyle \int_{0}^{\frac{\pi}{2}}\ln(\sin x)dx = -\frac{\pi}{2}\cdot \ln(2)+\bf{Imaginary\; quantity.}$$

But we have calculate only for real values.

So $$\displaystyle \int_{0}^{\frac{\pi}{2}}\ln(\sin x)dx = -\frac{\pi}{2}\cdot \ln(2)$$


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