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Ans -?ln2 Integration 0 to pi log(sin x)dx equals

Answers (1)

This can be calculated by using this technique:

\int_{0}^{\ pi}logsinxdx=\int_{0}^{\pi/2}logsinxdx+\int_{\frac{\pi}{2}}^{\pi}logsinxdx=I_1+I_2\\Let\ t=x-\frac{\pi}{2},dx=dt,\ x=\frac{\pi}{2}\Rightarrow t=0\\x=\pi\Rightarrow t=\frac{\pi}{2}\\\therefore I_2=\int_{0}^{\pi/2}logsinxdx\\2I=2\int_{0}^{\pi/2}logsinxdx\\ I= \int_{0}^{\pi/2}logsinxdx=\int_{0}^{\pi/2}logsin(x-\frac{\pi}{2})dx=\int_{0}^{\pi/2}logcosxdx\\2I=2\int_{0}^{\pi/2}logsin(\frac{2x}{2})dx=\int_{0}^{\pi/2}logsin2xdx-\int_{0}^{\pi/2}log2dx\\2x=u,dx=du/2\\x=0,u=0 and when x=\pi/2\ u=\pi\\I=\int_{0}^{\pi}logsin2xdx-\int_{0}^{\pi/2}log2dx\\\therefore \therefore i=-2\int_{0}^{\pi/2}log2dx=-\pi log2

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Safeer PP

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