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Question Number 75950 by turbo msup by abdo last updated on 21/Dec/19

$$give{\int }_0^{\frac{\pi }{2}}\frac{x}{sinx}dxatformofserie.$$

Commented by mathmax by abdo last updated on 22/Dec/19

$$letA={\int }_0^{\frac{\pi }{2}}\frac{x}{sinx}dxchangementtan\left(\frac{x}{2}\right)=tgive$$$$A=2{\int }_0^1\frac{arctan\left(t\right)}{\frac{2t}{1+t^2}}\frac{2dt}{1+t^2}=2{\int }_0^1\frac{arctan\left(t\right)}{t}dtwehave$$$$\frac{d}{dt}\left(arctant\right)=\frac{1}{1+t^2}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{t}^{2n}\Rightarrow$$$$arctan\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n\frac{t^{2n+1}}{2n+1}+c(c=0)\Rightarrow$$$$\frac{arctan\left(t\right)}{t}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n+1}{t}^{2n}\Rightarrow {\int }_0^1\frac{arctan\left(t\right)}{t}dt=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)}{\int }_0^1{t}^{2n}dt$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)}{\left[\frac{1}{2n+1}{t}^{2n+1}\right]}_0^1=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(2n+1)}^2}\Rightarrow$$$$A=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(2n+1)}^2}$$

Commented by mathmax by abdo last updated on 22/Dec/19

$$forgiveA=2\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(2n+1)}^2}$$

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