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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the following:$\int\limits_0^{\Large \pi} x \log \sin xdx$

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Toolbox:
  • $\int _0^af(x)dx=\int_0^af(a-x)dx$
  • $\int _0^2af(x)dx=2\int_0^af(2a-x)dx$
  • $2\sin x\cos x=\sin2x$
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}$$\log 2$
$\;\;=\large\frac{\pi^2}{2}$$\log\big(\large\frac{1}{2}\big)$
answered Apr 23, 2013 by sreemathi.v
 

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