Mathematics

Evaluate: $$I =\displaystyle \int \log \sin x dx$$


SOLUTION
$$=log\sin x\cdot \int dx - \int(\int dx)\cdot \dfrac{d}{dx}(log\sin x) dx + c$$

$$= log\sin x\cdot x - \int x \cdot \dfrac{1}{\sin x} \cdot \cos x dx + c$$

$$=xlog\sin x - \int x \cdot \dfrac{\cos x}{\sin x}dx + c$$

$$=xlog\sin x - \int x \cot x dx + c$$

$$=xlog\sin x - x \cdot \int \cot x dx - \int (\int \cot x dx) \dfrac{d}{dx}(x) dx +c +$$

$$=xlog \sin x - x \cdot log |\sin x| - \int log|\sin x|dx + c'$$

$$\because I = \int log |\sin x|$$

$$I = -I + c'$$

$$2I = c'$$

$$I = c''$$
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Subjective Medium Published on 17th 09, 2020
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