Mathematics

Evaluate the following:$\displaystyle \int_{0}^{\pi} xlog \,sin \,x \,dx$

SOLUTION
Let $I = \displaystyle \int_{0}^{\pi} xlog \,sin \,x \,dx$....(i)
$I = \displaystyle \int_{0}^{\pi} (\pi - x) log \,sin (\pi - x)dx$
$= \displaystyle \int_{0}^{\pi} (\pi -x) log \,sin \,x \,dx$ ...(ii)
$2I = \pi \displaystyle \int_{0}^{\pi} log \,sin \,x \,dx$ .....(iii)
$2I = 2\pi \displaystyle \int_{0}^{\pi/2} log \,sin \,x \,dx [\because \displaystyle \int{-}^{2a} f(x) = 2 \displaystyle \int_{0}^{a} f(x) \,dx]$
$I = \pi \displaystyle \int_{0}^{\pi/2} log \,sin \,x \,dx$ .....(iv)

Now, $I = \pi \displaystyle \int_{0}^{\pi/2} log \,sin (\pi / 2 - x)dx$ ....(v)
On adding Eqs. (iv) and (v), we get
$2I = \pi \displaystyle \int_{0}^{\pi/2} (log \,sin \,x + log \,cos \,x)dx$
$2I = \pi \displaystyle \int_{0}^{\pi/2} log \,sin \,x \,cos \,x \,dx$
$= \pi \displaystyle \int_{0}^{\pi / 22} log \dfrac{2 \,sin \,x \,cos \,x}{2}dx$
$2I = \pi \displaystyle \int_{0}^{\pi / 2} (log \,sin \,2x - log \,2)dx$
$2I = \pi \displaystyle \int_{0}^{\pi / 2} log \,sin \,2x dx - \pi \displaystyle \int_{0}^{\pi / 2} log \,2 \,dx$
Put $2x = t \Rightarrow dx = \dfrac{1}{2}dt$
As $x \rightarrow 0,$ then $t \rightarrow 0$
and $x \rightarrow \dfrac{\pi}{2},$ then $t \rightarrow \pi$
$\therefore 2I = \dfrac{\pi}{2} \displaystyle \int_{0}^{\pi} log \,sin \,t \,dt - \dfrac{\pi^2}{2} log \,2$
$\Rightarrow 2I = \dfrac{\pi}{2} \displaystyle \int_{0}^{\pi} log \,sin \,x \,dx - \dfrac{\pi^2}{2} log \,2$
$\Rightarrow 2I = \pi \displaystyle \int_{0}^{\pi/2} log \,sin \,x\, dx - \frac{\pi^2}{2} log \,2$
$\Rightarrow 2I = I - \frac{\pi^2}{2} log \,2$ (from Eqn (iv))
$\therefore I = -\dfrac{\pi^2}{2} log \,2 = \dfrac{\pi^2}{2} log \left ( \dfrac{1}{2} \right )$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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