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Question Number 197272 by Erico last updated on 12/Sep/23

$$\mathrm{How}\mathrm{to}\mathrm{calculate}\mathrm{this}\mathrm{integral}$$$${\underset{0}{\int }}^{\frac{\pi }{2}}\frac{\ln (1+sint)}{sint}dt$$

Answered by Mathspace last updated on 12/Sep/23

$$tan\left(\frac{t}{2}\right)=x\Rightarrow$$$$I={\int }_0^1\frac{ln(1+\frac{2x}{1+x^2})}{\frac{2x}{1+x^2}}\frac{2dx}{1+x^2}$$$$={\int }_0^1\frac{ln\left(\frac{1+x^2+2x}{1+x^2}\right)}{x}dx$$$$={\int }_0^1\frac{ln{(x+1)}^2-ln(1+x^2)}{x}dx$$$$=2{\int }_0^1\frac{ln(1+x)}{x}dx-{\int }_0^1\frac{ln(1+x^2)}{x}dx$$$$\frac{dx}{dx}ln(1+x)=\frac{1}{1+x}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^n$$$$\Rightarrow ln(1+x)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}{x}^{n+1}$$$$=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}\frac{x^n}{n}\Rightarrow$$$${\int }_0^1\frac{ln(1+x)}{x}dx=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{\int }_0^1{x}^{n-1}dx$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}=\eta \left(2\right)$$$$=(1-2^{1-2})\xi \left(2\right)=\frac{1}{2}.\frac{{\pi }^2}{6}=\frac{{\pi }^2}{12}$$$$ln(1+x^2)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{x}^{2n}\Rightarrow$$$${\int }_0^1\frac{ln(1+x^2)}{x}dx=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{\int }_0^1{x}^{2n-1}dx$$$$=\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}$$$$=\frac{1}{2}\frac{{\pi }^2}{12}=\frac{{\pi }^2}{24}\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}\frac{ln(1+sinx)}{sinx}dx$$$$=2.\frac{{\pi }^2}{12}-\frac{{\pi }^2}{24}=\frac{3{\pi }^2}{24}=\frac{{\pi }^2}{8}$$

Answered by universe last updated on 12/Sep/23

$$\mathit{I}={\int }_0^{\mathit{\pi }/2}\frac{\mathbf{log}(1+\mathit{\alpha }\mathbf{sin}\mathit{x})\mathit{d}\mathit{x}}{\mathbf{sin}\mathit{x}}$$$$\frac{\mathit{d}\mathit{I}}{\mathit{d}\mathit{\alpha }}={\int }_0^{\mathit{\pi }/2}\frac{1}{1+\mathit{\alpha }\mathbf{sin}\mathit{x}}\mathit{d}\mathit{x}$$$$\frac{\mathit{d}\mathit{I}}{\mathit{d}\mathit{\alpha }}={\int }_0^{\mathit{\pi }/2}\frac{{\mathbf{sec}}^2\mathit{x}/2}{1+{\mathbf{tan}}^2\mathit{x}/2+2\mathit{\alpha }\mathbf{tan}\mathit{x}/2}\mathit{d}\mathit{x}$$$$\frac{\mathit{d}\mathit{I}}{\mathit{d}\mathit{\alpha }}={\int }_0^1\frac{2\mathit{d}\mathit{y}}{{\mathit{y}}^2+2\mathit{\alpha }\mathit{y}+1}={\underset{0}{\int }}^1\frac{2\mathit{d}\mathit{y}}{{(\mathit{y}+\mathit{\alpha })}^2+(1-{\mathit{\alpha }}^2)}$$$$\frac{\mathit{d}\mathit{I}}{\mathit{d}\mathit{\alpha }}=\frac{2}{\sqrt{1-{\mathit{\alpha }}^2}}{\mathbf{tan}}^{-1}\frac{\mathit{y}+\mathit{\alpha }}{\sqrt{1-{\mathit{\alpha }}^2}}{\mid }_0^1$$$$\frac{\mathit{d}\mathit{I}}{\mathit{d}\mathit{\alpha }}=\frac{2}{\sqrt{1-{\mathit{\alpha }}^2}}[{\mathbf{tan}}^{-1}\frac{1-\mathit{\alpha }}{\sqrt{1-{\mathit{\alpha }}^2}}-{\mathbf{tan}}^{-1}\frac{\mathit{\alpha }}{\sqrt{1-{\mathit{\alpha }}^2}}]$$$$\frac{\mathit{d}\mathit{I}}{\mathit{d}\mathit{\alpha }}=\frac{2}{\sqrt{1-{\mathit{\alpha }}^2}}{\mathbf{tan}}^{-1}\left(\frac{1/\sqrt{1-{\mathit{\alpha }}^2}}{1+(1+\mathit{\alpha })\mathit{\alpha }/1-{\mathit{\alpha }}^2}\right)$$$$\int \frac{\mathit{d}\mathit{I}}{\mathit{d}\mathit{\alpha }}=\int \frac{2}{\sqrt{1-{\mathit{\alpha }}^2}}{\mathbf{tan}}^{-1}\sqrt{\frac{1-\mathit{\alpha }}{1+\mathit{\alpha }}}\mathit{d}\mathit{\alpha }$$$$\mathit{l}\mathit{e}\mathit{t}\mathit{\alpha }=\mathbf{cos}\mathit{\beta }\Rightarrow \mathit{d}\mathit{\alpha }=-\mathbf{sin}\mathit{\beta }\mathit{d}\mathit{\beta }$$$$\mathit{I}=-2\int {\mathbf{tan}}^{-1}\left(\sqrt{\frac{1-\mathbf{cos}\mathit{\beta }}{1+\mathbf{cos}\mathit{\beta }}}\right)\mathit{d}\mathit{\beta }$$$$\mathit{I}=-2\int \frac{\mathit{\beta }}{2}\mathit{d}\mathit{\beta }$$$$\mathit{I}=-\frac{{\mathit{\beta }}^2}{2}+\mathit{c}$$$$\mathit{I}=-\frac{{\left({\mathbf{cos}}^{-1}\mathit{\alpha }\right)}^2}{2}+\mathit{c}$$$$\mathit{l}\mathit{e}\mathit{t}\mathit{\alpha }=0\mathit{t}\mathit{h}\mathit{e}\mathit{n}\mathit{I}=0$$$$\mathit{c}=\frac{{\mathit{\pi }}^2}{8}$$$$\mathit{I}=\frac{{\mathit{\pi }}^2}{8}-\frac{{\left({\mathbf{cos}}^{-1}\mathit{\alpha }\right)}^2}{2}$$$$\mathit{n}\mathit{o}\mathit{w}\mathit{p}\mathit{u}\mathit{t}\mathit{\alpha }=1$$$$\mathit{I}=\frac{{\mathit{\pi }}^2}{8}$$

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