Question

$\int \limits_{0}^{\pi} \: x \: log \: sinx \: dx$
solve this .​

Answer 1

Step-by-step explanation:

$\sf I = \int\limits_{0}^{\pi}\:x\:log\:sinx\:dx$ ... ( 1 )

$\sf I = \int\limits_{0}^{\pi}\:( \pi - x ) \:log \:sin ( \pi - x ) dx$

$\therefore$$\sf I = \int\limits_{0}^{\pi}\:( \pi - x ) \:log \:sin x dx$ ... ( 2 )

$\sf adding \:equations\:(1)\:and\:( 2 ) </p><p></p><p></p><p>[tex]$

$\therefore$ $\sf 2I = \int\limits_{0}^{\pi}\:( \pi - x ) \:log \:sin x + x log sinx dx$

$\therefore$$\sf2I = \int\limits_{0}^{\pi}\: \pi \: log sinx dx$

$\therefore$$\sf2I =2\pi \int\limits_{0}^{\frac {\pi}{2}} log sinx dx$

$\therefore$$\sf I = \pi ( - \frac {\pi}{2} log 2 )$

$\therefore$$\sf I = - \frac {{\pi}^{2}}{2} log 2$

THANKS!!!

Answer 2

Answer:

-π²/2Log2

Step-by-step explanation:

Solution is in his given attachment !

____________________________________

Hope it's helpful