By using the properties of definite integrals, evaluate the integ…

Question

By using the properties of definite integrals, evaluate the integral in Exercise:
$\int^{\frac{\pi}{2}}_{0}\big(2\log\sin\text{x}-\log\sin2\text{x}\big)\text{dx}$

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Answer

$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\big(2\log\sin\text{x}-\log\sin2\text{x}\big)\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\big(\log\sin^{2}\text{x}-\log\sin2\text{x}\big)\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\log\bigg(\frac{\sin^{2}\text{x}}{\sin2\text{x}}\bigg)\text{dx}$ $=\int\limits_{0}^{\frac{\pi}{2}}\log\bigg(\frac{\sin^{2}\text{x}}{2\sin\text{x}\cos\text{x}}\bigg)\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\log\bigg(\frac{1}{2}\tan\text{x}\bigg)\text{dx}$ $\Rightarrow\ \ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\log\bigg(\frac{1}{2}\tan\bigg(\frac{\pi}{2}-\text{x}\bigg)\bigg)\text{dx}\ \bigg[\because\int\limits_{0}^{\text{a}}\text{f}\text{(x)}\text{dx}=\int\limits_{0}^{\text{a}}\text{f}\text{(a}-\text{x})\text{dx}=\bigg]$ $\Rightarrow\ \ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\log\bigg(\frac{1}{2}\cot\text{x}\bigg)\text{dx}$ Adding eq.(i) and (ii) $21=\int\limits_{0}^{\frac{\pi}{2}}\bigg[\log\bigg(\frac{1}{2}\tan\text{x}\bigg)+\log\bigg(\frac{1}{2}\cot\text{x}\bigg)\bigg]\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\bigg[\log\bigg(\frac{1}{2}\tan\text{x}\bigg)\bigg(\frac{1}{2}\cot\text{x}\bigg)\bigg]\text{dx}$ $\Rightarrow\ \ 21=\int\limits_{0}^{\frac{\pi}{2}}\bigg[\log\frac{1}{4}\bigg]\text{dx}=\log\frac{1}{4}\text{(x)}^{\frac{\pi}{2}}_{0}=\big(\log1-\log4\big)\frac{\pi}{2}=-\frac{\pi}{2}\log4$$\Rightarrow\ \ \text{I}=-\frac{\pi}{4}\log2^{2}=-\frac{2\pi}{4}\log2=-\frac{\pi}{2}\log2$