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Question Number 200254 by mnjuly1970 last updated on 16/Nov/23

$$$$$$calculate...$$$$\mathrm{\Omega }={\int }_{{\int }_0^{\frac{\pi }{2}}\ln \left(\tan \left(x\right)\right)dx}^{{\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\left(x\right)}{x^2}dx}\ln \left(\sin \left(x\right)\right)dx=?$$

Answered by Mathspace last updated on 16/Nov/23

$$itsatrikyintegral$$$$wehave{\int }_0^{\frac{\pi }{2}}ln\left(tanx\right)dx$$$$={\int }_0^{\frac{\pi }{2}}ln\left(\frac{sinx}{cosx}\right)dx$$$$={\int }_0^{\frac{\pi }{2}}ln\left(sinx\right)dx-{\int }_0^{\frac{\pi }{2}}ln\left(cosx\right)dx=0\left(equal\right)$$$${\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}x}{x^2}dx={[-\frac{1}{x}{sin}^{2}x]}_0^{\mathrm{\infty }}$$$$-{\int }_0^{\mathrm{\infty }}(-\frac{1}{x})2sinxcosxdx$$$$=0+{\int }_0^{\mathrm{\infty }}\frac{sin\left(2x\right)}{x}dx(2x=t)$$$$={\int }_0^{\mathrm{\infty }}\frac{sint}{\frac{t}{2}}\frac{dt}{2}={\int }_0^{\mathrm{\infty }}\frac{sint}{t}dt=\frac{\pi }{2}$$$$\Rightarrow I={\int }_0^{\frac{\pi }{2}}ln\left(sinx\right)dx$$$$=-\frac{\pi }{2}ln2$$

Commented by Calculusboy last updated on 16/Nov/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

Commented by mnjuly1970 last updated on 17/Nov/23

$$thanksalotsir$$

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