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Question Number 134708 by bramlexs22 last updated on 06/Mar/21

$$\mathcal{D}={\int }_0^{\pi /2}\sin {}^4\mathrm{x}\cos {}^5\mathrm{x}\mathrm{dx}$$

Answered by john_santu last updated on 06/Mar/21

$$\mathcal{D}={\int }_0^{\pi /2}\sin {}^{4}x{(1-\sin {}^{2}x)}^2\cos xdx$$$$letu=\sin x,\overline{)\begin{array}{c}u=1\left(upperlimit\right)\\ u=0\left(lowerlimit\right)\end{array}}$$$$\mathcal{D}={\int }_0^1{u}^4(u^4-2u^2+1)du$$$$\mathcal{D}={\int }_0^1(u^8-2u^6+u^4)du$$$$\mathcal{D}={[\frac{u^9}{9}-\frac{2u^7}{7}+\frac{u^5}{5}]}_0^1$$$$\mathcal{D}=\frac{1}{9}-\frac{2}{7}+\frac{1}{5}=\frac{35-90+63}{315}$$$$\mathcal{D}=\frac{8}{315}\bullet$$

Answered by Dwaipayan Shikari last updated on 06/Mar/21

$${\int }_0^{\frac{\pi }{2}}{sin}^{4}x{cos}^{5}xdx,Generally{\int }_0^{\frac{\pi }{2}}{sin}^{m}x{cos}^{n}xdx=\frac{\mathrm{\Gamma }\left(\frac{m+1}{2}\right)\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{2\mathrm{\Gamma }(\frac{m+n}{2}+1)}$$$$=\frac{\mathrm{\Gamma }\left(\frac{5}{2}\right)\mathrm{\Gamma }\left(3\right)}{2\mathrm{\Gamma }\left(\frac{11}{2}\right)}=\frac{\frac{3}{4}\sqrt{\pi }.2}{2.\frac{9}{2}.\frac{7}{2}.\frac{5}{2}.\frac{3}{4}\sqrt{\pi }}=\frac{8}{315}$$

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