Mathematics

Evaluate; $$\displaystyle\int _ { 0 } ^ { \pi/2 } \log \sin  { 2 } x d x$$


SOLUTION

$$\\I=\int_{0}^{(\frac{\pi}{2})}log sin2x dx\\Let 2x=t\\2dx=dt\\\therefore dx=(\frac{1}{2})dt\\and x\rightarrow(\frac{\pi}{2}), t\rightarrow\pi\\x\rightarrow 0, t\rightarrow 0\\I=\int_{0}^{\pi}logsin(t)(\frac{1}{2})dt\\=(\frac{1}{2})\int_{0}^{\pi}logsin(\pi-t)dt\\=(\frac{1}{2})\int_{0}^{\pi}logsin(\pi-t)dt\\=(\frac{1}{2})\int_{0}^{\pi}logsin\>t\>dt\\=(\frac{1}{2})\times2\int_{0}^{(\frac{\pi}{2})}log\>sint\>dt\\=-(\frac{\pi}{2})log2(standard\>result)$$

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Subjective Medium Published on 17th 09, 2020
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