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Question Number 75960 by Tony Lin last updated on 21/Dec/19

$$provethat{\int }_0^{\frac{\pi }{2}}ln\left(sinx\right)dx=-\frac{\pi }{2}ln2$$

Commented by Kunal12588 last updated on 21/Dec/19

$$nowsolve{\int }_0^{\frac{\pi }{2}}ln\left(cosx\right)dx$$

Commented by Tony Lin last updated on 23/Dec/19

$${\int }_0^{\frac{\pi }{2}}ln\left(sinx\right)dx$$$$let\theta =\frac{\pi }{2}-x,d\theta =-dx$$$$-{\int }_{\frac{\pi }{2}}^{0}ln\left[sin(\frac{\pi }{2}-x)\right]d\theta$$$$={\int }_0^{\frac{\pi }{2}}ln\left(cosx\right)dx$$

Answered by Kunal12588 last updated on 21/Dec/19

$$I={\int }_0^{\frac{\pi }{2}}ln\left(sinx\right)dx$$$$\Rightarrow I={\int }_0^{\frac{\pi }{2}}ln\left(cosx\right)dx$$$$2I={\int }_0^{\frac{\pi }{2}}ln\left(sinxcosx\right)dx$$$$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}[ln\left(sin2x\right)-ln\left(2\right)]dx$$$$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}ln\left(sin2x\right)dx-{\left[xln2\right]}_0^{\frac{\pi }{2}}$$$$lett=2x\Rightarrow dx=\frac{1}{2}dt$$$$x\to 0\Rightarrow t\to 0$$$$x\to \frac{\pi }{2}\Rightarrow t\to \pi$$$$2I=\frac{1}{2}{\int }_0^{\pi }ln\left(sint\right)dt-\frac{\pi }{2}ln2$$$$\Rightarrow 2I=\frac{1}{2}\times 2{\int }_0^{\frac{\pi }{2}}ln\left(sinx\right)dx-\frac{\pi }{2}ln2$$$$\Rightarrow 2I=I-\frac{\pi }{2}ln2$$$$\Rightarrow I=-\frac{\pi }{2}ln2proved$$

Commented by Tony Lin last updated on 23/Dec/19

$$thankssir$$

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