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Question Number 129490 by 676597498 last updated on 16/Jan/21

$${\int }_0^{\frac{\pi }{2}}\mathrm{x}\sqrt{\mathrm{tanx}}\mathrm{dx}$$

Answered by mnjuly1970 last updated on 16/Jan/21

$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{2}}x.{sin}^{\frac{1}{2}}\left(x\right).{cos}^{\frac{1}{2}}\left(x\right)dx$$$$=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}{sin}^{\frac{1}{2}}\left(x\right).{cos}^{\frac{1}{2}}\left(x\right)-\mathrm{\Omega }$$$$2\mathrm{\Omega }=\frac{\pi }{4}[2{\int }_0^{\frac{\pi }{2}}{sin}^{\frac{1}{2}}\left(x\right).{cos}^{\frac{1}{2}}\left(x\right)dx]$$$$=\frac{\pi }{4}\left(\beta (\frac{3}{4},\frac{3}{4})\right)=\frac{\pi }{4}\ast \frac{{\mathrm{\Gamma }}^2\left(\frac{3}{4}\right)}{\mathrm{\Gamma }\left(\frac{3}{2}\right)}$$$$\therefore \mathrm{\Omega }=\frac{\pi }{8}\ast \frac{{\mathrm{\Gamma }}^2\left(\frac{3}{4}\right)}{\frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)}=\frac{\pi }{4\sqrt{\pi }}\ast \mathrm{\Gamma }{}^2\left(\frac{3}{4}\right)$$

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