int_0^pi x log(sin x) , dx = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int_0^\pi x \log(\sin x) \, dx$ ..... $(i)$ Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get: $I = \int_0^\pi (\p

Vedclass · Accepted Answer

$\frac{\pi^2}{2} \log(\frac{1}{2})$

Vedclass · Answer

(B) Let $I = \int_0^\pi x \log(\sin x) \, dx$ ..... $(i)$ Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get: $I = \int_0^\pi (\pi - x) \log(\sin(\pi - x)) \, dx = \int_0^\pi (\pi - x) \log(\sin x) \, dx$ ..... $(ii)$ Adding $(i)$ and $(ii)$: $2I = \int_0^\pi (x + \pi - x) \log(\sin x) \, dx = \pi \int_0^\pi \log(\sin x) \, dx$ Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$: $2I = 2\pi \int_0^{\pi/2} \log(\sin x) \, dx$ $I = \pi \int_0^{\pi/2} \log(\sin x) \, dx$ We know that $\int_0^{\pi/2} \log(\sin x) \, dx = -\frac{\pi}{2} \log 2 = \frac{\pi}{2} \log(\frac{1}{2})$. Therefore,$I = \pi \left( \frac{\pi}{2} \log(\frac{1}{2}) \right) = \frac{\pi^2}{2} \log(\frac{1}{2})$.

$\int_0^\pi x \log(\sin x) \, dx = $

Similar Questions

Let $I(R) = \int_0^R e^{-R \sin x} dx$, where $R > 0$. Then,

$\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$ is equal to :

If $f(x) = \int\limits_0^x {2(\cos^2 3t + 3\sin^2 3t)dt}$,then $f(x + \pi)$ is equal to

The value of $\sum \limits_{n=0}^{1947} \frac{1}{2^n+\sqrt{2^{1947}}}$ is equal to

$\int_0^{\pi /2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module