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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evaluate the following integrals:
$\int\limits^{\pi}_0\text{x}\log\sin\text{x}\text{ dx}$
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Answer

$\int\limits^{\pi}_0\text{x}\log\sin\text{x}\text{ dx}$
Let $\text{I}=\int\limits^{\pi}_0\text{x}\log\sin\text{x}\text{ dx}\ ...(\text{i})$
$\text{I}=\int\limits^{\pi}_0(\pi-\text{x})\log\sin(\pi-\text{x})\text{dx}$
$\text{I}=\int\limits^{\pi}_0(\pi-\text{x})\log(\sin\text{x})\text{dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\pi\int\limits^{\pi}_0\log\sin\text{x}\text{ dx}$
$=2\pi\int\limits^{\frac{\pi}{2}}_0\log\sin\text{x}\text{ dx}$
$\text{I}=\pi\int\limits^{\frac{\pi}{2}}_0\log\sin\text{x}\text{ dx}\ ....(\text{iii})$
Let $\int\limits^{\frac{\pi}{2}}_0\log\sin\text{x}\text{ dx}=\text{I}_2$
$\text{I}_2=\int\limits^{\frac{\pi}{2}}_0\log\sin\Big(\frac{\pi}{2}-\text{x}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\cos\text{x dx}$
$2\text{I}_2=\int\limits^{\frac{\pi}{2}}_0(\log\sin\text{x}+\log\cos\text{x})\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log(\sin\text{x}\cos\text{x})\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log(\sin2\text{x})\text{dx}-\int\limits^{\frac{\pi}{2}}_0\log2\text{ dx}$
Let $2\text{x}=\text{t}$
$2\text{dx}=\text{dt}$
When, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=0\Rightarrow\text{t}=\pi$
$2\text{I}_2=\frac{1}{2}\int\limits^{{\pi}}_0\log(\sin\text{t})\text{dt}-\frac{\pi}{2} \log2 $
$2\text{I}_2=\frac{2}{2}\int\limits^{{\pi}}_0\log(\sin\text{t})\text{dt}-\frac{\pi}{2} \log2 $
$2\text{I}_2=\text{I}_2-\frac{\pi}{2} \log2 $
$\text{I}_2=-\frac{\pi}{2}\log2$
From (iii),
$\text{I}=\pi\int\limits^{{\pi}}_0\log\sin\text{x}\text{dx}=\pi\text{I}_2$
$\text{I}=\pi\Big(-\frac{\pi}{2}\log2\Big)$
$\text{I}=\frac{-\pi^2\log2}{2}$

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