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Question Number 140628 by mohammad17 last updated on 10/May/21

$$withoutspecialfunctionfind$$$$$$$${\int }_0^{\frac{\pi }{10}}\sqrt{tanx}dx?$$$$$$

Answered by Dwaipayan Shikari last updated on 10/May/21

$$\int \sqrt{tanx}dxtanx=t^2$$$$=\int \frac{2t^2}{1+t^4}dt=2\int \frac{1}{(t^2+\frac{1}{t^2})}dt=\int \frac{1-\frac{1}{t^2}}{{(t+\frac{1}{t})}^2-2}+\frac{1+\frac{1}{t^2}}{{(t-\frac{1}{t})}^2+2}dt$$$$=\frac{1}{2\sqrt{2}}log\left(\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right)+\frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+C$$$$=\frac{1}{2\sqrt{2}}log\left(\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right)+\frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{t^2-1}{\sqrt{2t}}\right)+C$$$${\int }_0^{\frac{\pi }{10}}\sqrt{tanx}dx=\frac{1}{2\sqrt{2}}log\left(\frac{tan\frac{\pi }{10}-\sqrt{2tan\frac{\pi }{10}}+1}{tan\frac{\pi }{10}+\sqrt{2tan\frac{\pi }{10}}+1}\right)+\frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{tan\frac{\pi }{10}-1}{\sqrt{2tan\frac{\pi }{10}}}\right)+\frac{\pi }{2\sqrt{2}}$$$$tan\frac{\pi }{10}=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}$$$$=\frac{1}{\sqrt{2}}log(\frac{\sqrt{5}-1}{10+2\sqrt{5}}-\sqrt{\frac{\sqrt{5}-1}{5+\sqrt{5}}}+1)-\frac{1}{2\sqrt{2}}log\left(\frac{8}{5+\sqrt{5}}\right)+\frac{1}{\sqrt{2}}{tan}^{-1}\left(\frac{\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}-1}{\sqrt{\frac{\sqrt{5}-1}{5+\sqrt{5}}}}\right)+\frac{\pi }{2\sqrt{2}}$$

Answered by Dwaipayan Shikari last updated on 10/May/21

$${\int }_0^a{sin}^{\alpha }\left(x\right){cos}^{\beta }\left(x\right)dx$$$$={\int }_0^{sin\left(a\right)}\frac{t^{\alpha }}{\sqrt{1-t^2}}{\left(\sqrt{1-t^2}\right)}^{\beta }dt$$$$={\int }_0^{sin\left(a\right)}\frac{t^{\alpha }}{{(1-t^2)}^{\frac{1-\beta }{2}}}dt=\sum _{n⩾0}{\int }_0^{sin\left(a\right)}\frac{{\left(\frac{1-\beta }{2}\right)}_n}{n!}{t}^{\alpha +2n}dt$$$$={sin}^{\alpha +1}\left(a\right)\sum _{n⩾0}\frac{{\left(\frac{1-\beta }{2}\right)}_n}{n!(\alpha +2n+1)}{\left({sin}^2\left(a\right)\right)}^{2n}$$$$=\frac{1}{\alpha +1}{sin}^{\alpha +1}\left(a\right)\sum _{n⩾0}\frac{{\left(\frac{1-\beta }{2}\right)}_n{\left(\frac{\alpha +1}{2}\right)}_n}{n!{\left(\frac{\alpha +3}{2}\right)}_n}{\left({sin}^2\left(a\right)\right)}^{2n}$$$$=\frac{{sin}^{\alpha +1}\left(a\right)}{\alpha +1}{}_2{F}_1(\frac{1-\beta }{2},\frac{\alpha +1}{2};\frac{\alpha +3}{2};{sin}^2\left(a\right))$$$${\int }_0^{\frac{\pi }{10}}\sqrt{tanx}dx=\frac{{sin}^{3/2}\left(\frac{\pi }{10}\right)}{\frac{1}{2}+1}{}_2{F}_1(\frac{3}{4},\frac{3}{4};\frac{7}{4};\frac{3-\sqrt{5}}{8})$$$$=\frac{2}{3}{\left(\frac{3-\sqrt{5}}{8}\right)}^{3/2}{}_2{F}_1(\frac{3}{4};\frac{3}{4};\frac{7}{4};\frac{3-\sqrt{5}}{8})$$

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