Question
$\int\limits_{0}^{\pi}\text{x}\log\sin\text{x dx}$
✓
Answer
Let $\text{I}=\int\limits_{0}^{\pi}\text{x}\log\sin\text{x dx}\ \ \dots(\text{i})$
$\Rightarrow\ \text{I}=\int\limits^{\pi}_{0}(\pi-\text{x})\log\sin(\pi-\text{x})\text{dx}$ $\bigg[\because\ \int\limits^{\text{b}}_{\text{a}}\text{f(x)dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\bigg]$
$\Rightarrow\ \text{I}=\int\limits^{\pi}_{0}(\pi-\text{x})\log\sin\text{x}\text{ dx}\ \ \dots(\text{ii})$
Adding (i) and (ii), we get
$2\text{I}=\pi\int^{\pi}_{0}\log\sin\text{x}\text{ dx} \ \ \dots(\text{iii})$
$2\text{I}=2\pi\int\limits^{\frac{\pi}{2}}_{0}\log\sin\text{x}\text{ dx} $
$ \bigg[\because\ \int\limits^{2\text{a}}_{0}\text{f}(\text{x})\text{dx}=2\int\limits^{\text{a}}_{0}\text{f}(\text{x})\text{dx if f}(2\text{a}-\text{x})=\text{f(x)}\bigg] $
$\Rightarrow\ \text{I}=\pi\int\limits^{\frac{\pi}{2}}_{0}\log\sin\text{x dx} \ \ \dots(\text{iv})$
$\Rightarrow\ \text{I}=\pi\int\limits^{\frac{\pi}{2}}_{0}\log\sin\Big(\frac{\pi}{2}-\text{x}\Big)\text{ dx} \ \ \dots(\text{v})$
Adding (iv) and (v), we get
$2\text{I}=\pi\int\limits^{\frac{\pi}{2}}_{0}(\log\sin\text{x}+\log\cos\text{x})\text{ dx} $
$=\pi\int\limits^{\frac{\pi}{2}}_{0}\log\sin\text{x}\cos\text{x}\text{ dx}$
$=\pi\int\limits^{\frac{\pi}{2}}_{0}\log\frac{2\sin\text{x}\cos\text{x}}{2}\text{ dx} $
$=\pi\int\limits^{\frac{\pi}{2}}_{0}(\log\sin2\text{x}-\log2)\text{ dx} $
$=\pi\int\limits^{\frac{\pi}{2}}_{0}\log\sin2\text{x dx}-\pi\int\limits^{\frac{\pi}{2}}_{0}\log2\text{ dx} $
Put $2\text{x}=\text{t}\Rightarrow\ \text{dx}=\frac{1}{2}\text{dt}$
$\therefore\ 2\text{I}=\frac{\pi}{2}\int\limits^{\pi}_{2}\log\sin\text{t dt}-\frac{\pi^2}{2}\log2$
$\Rightarrow\ \ 2\text{I}=\frac{\pi}{2}\int\limits^{\pi}_{2}\log\sin\text{x dx}-\frac{\pi^2}{2}\log2$
$\Rightarrow\ 2\text{I}=\text{I}-\frac{\pi^2}{2}\log2 $ [from Eq. (iii)]
$\therefore\ \text{I}=-\frac{\pi^2}{2}\log2 $
$\Rightarrow\ \text{I}=\int\limits^{\pi}_{0}(\pi-\text{x})\log\sin(\pi-\text{x})\text{dx}$ $\bigg[\because\ \int\limits^{\text{b}}_{\text{a}}\text{f(x)dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\bigg]$
$\Rightarrow\ \text{I}=\int\limits^{\pi}_{0}(\pi-\text{x})\log\sin\text{x}\text{ dx}\ \ \dots(\text{ii})$
Adding (i) and (ii), we get
$2\text{I}=\pi\int^{\pi}_{0}\log\sin\text{x}\text{ dx} \ \ \dots(\text{iii})$
$2\text{I}=2\pi\int\limits^{\frac{\pi}{2}}_{0}\log\sin\text{x}\text{ dx} $
$ \bigg[\because\ \int\limits^{2\text{a}}_{0}\text{f}(\text{x})\text{dx}=2\int\limits^{\text{a}}_{0}\text{f}(\text{x})\text{dx if f}(2\text{a}-\text{x})=\text{f(x)}\bigg] $
$\Rightarrow\ \text{I}=\pi\int\limits^{\frac{\pi}{2}}_{0}\log\sin\text{x dx} \ \ \dots(\text{iv})$
$\Rightarrow\ \text{I}=\pi\int\limits^{\frac{\pi}{2}}_{0}\log\sin\Big(\frac{\pi}{2}-\text{x}\Big)\text{ dx} \ \ \dots(\text{v})$
Adding (iv) and (v), we get
$2\text{I}=\pi\int\limits^{\frac{\pi}{2}}_{0}(\log\sin\text{x}+\log\cos\text{x})\text{ dx} $
$=\pi\int\limits^{\frac{\pi}{2}}_{0}\log\sin\text{x}\cos\text{x}\text{ dx}$
$=\pi\int\limits^{\frac{\pi}{2}}_{0}\log\frac{2\sin\text{x}\cos\text{x}}{2}\text{ dx} $
$=\pi\int\limits^{\frac{\pi}{2}}_{0}(\log\sin2\text{x}-\log2)\text{ dx} $
$=\pi\int\limits^{\frac{\pi}{2}}_{0}\log\sin2\text{x dx}-\pi\int\limits^{\frac{\pi}{2}}_{0}\log2\text{ dx} $
Put $2\text{x}=\text{t}\Rightarrow\ \text{dx}=\frac{1}{2}\text{dt}$
$\therefore\ 2\text{I}=\frac{\pi}{2}\int\limits^{\pi}_{2}\log\sin\text{t dt}-\frac{\pi^2}{2}\log2$
$\Rightarrow\ \ 2\text{I}=\frac{\pi}{2}\int\limits^{\pi}_{2}\log\sin\text{x dx}-\frac{\pi^2}{2}\log2$
$\Rightarrow\ 2\text{I}=\text{I}-\frac{\pi^2}{2}\log2 $ [from Eq. (iii)]
$\therefore\ \text{I}=-\frac{\pi^2}{2}\log2 $
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