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Question Number 32480 by prof Abdo imad last updated on 25/Mar/18

$$find{\int }_0^{\alpha }\sqrt{tanx}dxwith0<\alpha <\frac{\pi }{2}.$$

Commented byprof Abdo imad last updated on 14/Apr/18

$$letputI={\int }_0^{\alpha }\sqrt{tanx}dx.changement\sqrt{tanx}=t$$ $$givex=arctan\left(t^2\right)and$$ $$I={\int }_0^{\sqrt{tan\alpha }}t.\frac{2tdt}{1+t^4}={\int }_0^{\sqrt{tan\alpha }}\frac{2t^2}{1+t^4}dt$$ $$=\frac{1}{2}{\int }_0^{\sqrt{tan\alpha }}\frac{1}{t}\frac{4t^3}{1+t^4}dtbypartsu=\frac{1}{t}and$$ $$v^{{}^{\prime }}=\frac{4t^3}{1+t^4}\Rightarrow I={\left[\frac{1}{t}ln(1+t^4)\right]}_0^{\sqrt{tan\alpha }}$$ $$-{\int }_0^{\sqrt{tan\alpha }}\frac{-1}{t^2}ln(1+t^4)dt$$ $$=\frac{1}{\sqrt{tan\alpha }}ln(1+{tan}^2\alpha )+{\int }_0^{\sqrt{tan\alpha }}\frac{ln(1+t^4)}{t^2}dt$$ $$if0<\alpha <\frac{\pi }{4}\Rightarrow 0<\sqrt{tan\alpha }<1soletdevelopp$$ $$ln(1+t^4)wehaveln{(1+u)}^{{}^{\prime }}=\mathrm{\Sigma }{(-1)}^n{u}^n$$ $$ln(1+u)=\mathrm{\Sigma }{}_{n⩾0}\frac{{(-1)}^n{u}^{n+1}}{n+1}=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}\frac{u^n}{n}$$ $$ln(1+x^4)=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}\frac{x^{4n}}{n}\Rightarrow$$ $$\frac{ln(1+t^4)}{t^2}=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}\frac{t^{4n-2}}{n}\Rightarrow$$ $${\int }_0^{\sqrt{tan\alpha }}\frac{ln(1+t^4)}{t^2}dt=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{\int }_0^{\sqrt{tan\alpha }}{t}^{4n-2}dt$$ $$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}\frac{1}{4n-1}{\left(tan\alpha \right)}^{2n-1}letput$$ $$S\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}\frac{1}{4n-1}{x}^{2n-1}letfindS\left(x\right)$$ $$\frac{1}{4}S\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{(4n-1)\left(4n\right)}{x}^{2n-1}$$ $$=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}(\frac{1}{4n-1}-\frac{1}{4n})x^{2n-1}$$ $$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{4n-1}{x}^{2n-1}-\frac{1}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{x}^{2n-1}$$ $$...becontinued...$$

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