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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

$I_1= \int \limits _0^{\pi} x \log \sin x dx;\; I_2=\int \limits_0^{\pi} \log \sin x dx . $ find $\large\frac{I_1}{I_2}$=?

\[\begin {array} {1 1} (a)\;\frac{\pi}{2} \\ (b)\;\frac{\pi}{4} \\ (c)\;\frac{\pi}{6} \\ (d)\;\frac{\pi}{8} \end {array}\]

1 Answer

$f(x)= \log \sin x $
$f(\pi-x)= \log \sin x $
$f(x)=f( \pi-x) $
=> $I_1= \int\limits_0^{\pi} x \log \sin x dx $
$\qquad= \large\frac{z}{2}\int\limits_0^{z} $$ \log \sin x dx $
$I_1= \large\frac{z}{2} $$I_2$
$\large\frac{I_1}{I_2}= \frac{\pi}{2}$
answered Dec 13, 2013 by meena.p
 
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