Evaluate the following integrals: ^ 2 _0 (2 - 2x )dx

Question

Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\big(2\log\cos\text{x}-\log\sin2\text{x}\big)\text{dx}$

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Answer

We have,
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\big(2\log\cos\text{x}-\log\sin2\text{x}\big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\big(\log\cos^2\text{x}-\log\sin2\text{x}\big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\frac{\cos^2\text{x}}{\sin\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\frac{\cos^2\text{x}}{2\sin\text{x}\cdot\cos\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\frac{\cos\text{x}}{2\sin\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\big(\log\cos\text{x}-\log\sin\text{x}-\log2\big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\cos\text{x dx}-\int\limits^{\frac{\pi}{2}}_0\log\sin\text{x dx}-\int\limits^{\frac{\pi}{2}}_0\log2$
We know that
$\int\limits^{\frac{\pi}{2}}_0\log\cos\text{x dx}=\int\limits^{\frac{\pi}{2}}_0\log\sin\text{x dx}\ ...(\text{i})$
Hence from equation (i)
$\text{I}=-\int\limits^{\frac{\pi}{2}}_0\log2=-\frac{\pi}{2}\log2$

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