First slide

Evaluation of definite integrals

Question

$Evaluate \int_{0}^{π} x \log \sin x d x$

Difficult

A

$\frac{1}{2} π^{2} \log (1 / 2)$
B

$2 π^{2} \log (1 / 2)$
C

$π^{2} \log (1 / 2)$
D

$π^{2} \log (2)$

Solution

$Let I = \int_{0}^{π} x \log \sin x d x$ --------(1)
$I = \int_{0}^{π} (π - x) \log \sin (π - x) d x = \int_{0}^{π} (π - x) \log \sin x d x$ ---------(2)
$Adding equations (1) and (2), we get 2 I = π \int_{0}^{π} \log \sin x d x$
$or 2 I = 2 π \int_{0}^{π / 2} \log \sin x d x$
$I = π \int_{0}^{π / 2} \log \sin x d x - - - - (3) I = π \int_{0}^{π / 2} \log \sin (\frac{π}{2} - x) d x = π \int_{0}^{π / 2} \log c o s x d x - - - - (4)$
(3) + (4)
$2 I = π \int_{0}^{π / 2} (\log \sin x + logcos x) d x$
$(t a k e logsin x \cos x = \log \frac{2 \sin x \cos x}{2})$
$= π \int_{0}^{π / 2} \log \sin 2 x d x - π \int_{0}^{π / 2} \log 2 d x$
put 2x=t, dx= $\frac{d t}{2}$ UL= $π$ , LL=0
$2 I$ = $\frac{π}{2} \int_{0}^{π} \log \sin t d t - π \log 2 (\frac{π}{2})$
$2 I = π \int_{0}^{\frac{π}{2}} \log \sin tdt - \frac{π^{2}}{2} \log 2$
$2 I = I - \frac{π^{2}}{2} \log 2$
$I = - \frac{π^{2}}{2} \log 2$
$= \frac{π^{2}}{2} \log (1 / 2)$

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