Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

Evaluate the following:$\int\limits_0^{\Large \pi} x \log \sin xdx$

Can you answer this question?

1 Answer

0 votes
  • $\int _0^af(x)dx=\int_0^af(a-x)dx$
  • $\int _0^2af(x)dx=2\int_0^af(2a-x)dx$
  • $2\sin x\cos x=\sin2x$
Step 1:
Let $I=\int_0^\pi x\log \sin x dx$-----(1)
By applying the property $\int _0^af(x)dx=\int_0^af(a-x)dx$
Let $I=\int_0^\pi(\pi-x)\log\sin (\pi-x)dx$
But $\sin(\pi-x)=\sin x$
$\;\;=\int_0^\pi(\pi-x)\log\sin x dx$------(2)
Adding equ(1) & (2) we get,
$2I=\int _0^\pi xlog \sin xdx+\int_0^\pi(\pi-x)\log\sin xdx.$
$\;\;=\int_0^\pi [x\log \sin xdx+\pi \log \sin xdx-x\log \sin x dx]$
$\;\;=\pi \int \log \sin x dx$.
Let $I=\pi\int_0^\pi \log \sin x dx$------(3)
Again apply the property $\int_0^{2a}f(x)dx=2\int_0^af(2a-x)dx.$
$I=2\pi\int_0^{\large\frac{\pi}{2}} \log \sin x dx$
By applying the property $\int_0^af(x)dx=2\int_0^af(a-x)dx.$
$I=2\pi\int_0^{\large\frac{\pi}{2}} \log \sin (\large\frac{\pi}{2}\normalsize-x) dx$------(4)
But $\sin(\large\frac{\pi}{2}\normalsize-x)=\cos x$
Hence $I=\pi\int_0^{\large\frac{\pi}{2}}\log \cos x dx$------(5)
Step 2:
Now adding equ(4) &(5) we get
$2I=2\pi\begin{bmatrix}\int_0^{\large\frac{\pi}{2}}\log \sin xdx+\int_0^{\large\frac{\pi}{2}}\log \cos xdx\end{bmatrix}$
$\;\;=2\pi[\int_0^{\large\frac{\pi}{2}}\log (\sin x\cos x)dx]$
But $\sin x\cos x=\large\frac{1}{2}\normalsize\sin 2x$
$I=\pi \int_0^{\large\frac{\pi}{2}}\log \large\frac{\sin 2x}{2}\normalsize dx$
$\;\;=\pi\begin{bmatrix}\int_0^{\large\frac{\pi}{2}}\log 2xdx-\int_0^{\large\frac{\pi}{2}}\log 2dx\end{bmatrix}$
But $\int_0^{\large\frac{\pi}{2}}\log 2dx=[\log 2.x]=\large\frac{\pi}{2}\normalsize\log 2$
$I=\pi\begin{bmatrix}\int_0^{\large\frac{\pi}{2}}\log \sin2xdx-\large\frac{\pi}{2}\normalsize\log 2\end{bmatrix}$
Step 3:
Now let $I_1=\int_0^{\large\frac{\pi}{2}}log \sin 2xdx.$
By putting $2x=t\Rightarrow 2dx=dt$ or $dx=\large\frac{dt}{2}$
$I_1=\int_0^\pi\log \sin t.\large\frac{dt}{2}.$
By applying the property $\int_0^{2a}f(x)dx=\int_0^af(a-x)dx.$
$I_1=\large\frac{2\pi}{2}\int_0^{\large\frac{\pi}{2}}\normalsize\log \sin t. dt.$
$\;\;=\pi\int_0^{\large\frac{\pi}{2}}\log x dx=I$
$2I=[I-\large\frac{\pi^2}{2}\normalsize\log 2]$
$I=\large\frac{-\pi^2}{2}$$\log 2$
answered Apr 23, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App