How do you evaluate the integral $$\int_{0}^{\frac{\pi}{2}} \log(\sin x)\ dx\ \text{?}$$
Asked
Active
Viewed 565 times
-3
-
do you want to know this integral $$\int_{0}^{\pi{2}}\log(\sin(x))dx$$? – Dr. Sonnhard Graubner Sep 27 '15 at 08:23
-
Welcome to MathSE. When you pose a question here, you should include your own thoughts on the problem. What have you attempted? Where are you stuck? – N. F. Taussig Sep 27 '15 at 08:25
-
Yes please answer – Animesh Sep 27 '15 at 08:26
-
i think this integral was always solved in this forum – Dr. Sonnhard Graubner Sep 27 '15 at 08:30
-
I simply integrated log sin x DX as 1/sinx and put limits – Animesh Sep 27 '15 at 08:31
-
@Animesh, why did you do that? $\log(\sin x) \ne \frac{1}{\sin x}$ – Galc127 Sep 27 '15 at 08:32
-
Please post your and. – Animesh Sep 27 '15 at 08:35
2 Answers
2
HINT: notice we have $$I=\int_{0}^{\pi/2}\log(\sin x)\ dx\tag 1$$
Using property of definite integral, we get
$$I=\int_{0}^{\pi/2}\log\left(\sin\left(\frac{\pi}{2}-x\right) \right)\ dx$$ $$I=\int_{0}^{\pi/2}\log\left(\cos x\right)\ dx\tag 2$$
Adding (1) & (2), we get $$I+I=\int_{0}^{\pi/2}\log\left(\sin x\cos x\right)\ dx$$
$$2I=\int_{0}^{\pi/2}\log\left(\frac{2\sin x\cos x}{2}\right)\ dx$$
$$2I=\int_{0}^{\pi/2}\log\left(\sin 2x\right)\ dx-\int_{0}^{\pi/2}\log(2)\ dx$$
Hopefully, You can solve further
Harish Chandra Rajpoot
- 38,565
2
Hint
Show that $$\int_0^{\pi/2}\ln(\sin(x))dx=\int_0^{\pi/2}\ln(\cos(x))dx.$$
Then $$\int_0^{\pi/2}\ln(\sin(x))dx=\frac{1}{2}\left(\int_0^{\pi/2}\ln(\sin(x))dx+\int_0^{\pi/2}\ln(\cos(x))dx\right)=\frac{1}{2}\int_0^{\pi/2}\ln\left(\frac{\sin(2x)}{2}\right)dx\underset{u=2x}{=}\frac{1}{4}\int_0^{\pi}\ln(\sin(u))du-\frac{\pi\ln 2}{4}=...$$
Surb
- 57,262
- 11
- 68
- 119