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Question Number 145487 by Mrsof last updated on 05/Jul/21

Commented by Mrsof last updated on 05/Jul/21

$$helpmesir$$

Answered by Ar Brandon last updated on 05/Jul/21

$$\mathrm{I}={\int }_0^{\pi }{\tan }^5\left(\mathrm{x}\right){\sec }^3\left(\mathrm{x}\right)\mathrm{dx}$$$$={\int }_0^{\pi }{\tan }^4\left(\mathrm{x}\right){\sec }^2\left(\mathrm{x}\right)\cdot \sec \left(\mathrm{x}\right)\tan \left(\mathrm{x}\right)\mathrm{dx}$$$$={\int }_0^{\pi }{({\sec }^2\left(\mathrm{x}\right)-1)}^2{\sec }^2\left(\mathrm{x}\right)\mathrm{d}\left(\sec \left(\mathrm{x}\right)\right)$$$$={\int }_0^{\pi }({\sec }^6\left(\mathrm{x}\right)-{2\sec }^4\left(\mathrm{x}\right)+{\sec }^2\left(\mathrm{x}\right))\mathrm{d}\left(\sec \left(\mathrm{x}\right)\right)$$$$={[\frac{{\sec }^7\left(\mathrm{x}\right)}{7}-\frac{{2\sec }^5\left(\mathrm{x}\right)}{5}+\frac{{\sec }^3\left(\mathrm{x}\right)}{3}]}_0^{\pi }$$$$=(-\frac{1}{7}+\frac{2}{5}-\frac{1}{3})-(\frac{1}{7}-\frac{2}{5}+\frac{1}{3})=-\frac{16}{105}$$$$\mid \mathrm{I}\mid =\mid -\frac{16}{105}\mid =\frac{16}{105}$$

Commented by Ar Brandon last updated on 05/Jul/21

$$\mathrm{I}={\int }_0^{\pi }{\tan }^5\left(\mathrm{x}\right){\sec }^3\left(\mathrm{x}\right)\mathrm{dx}$$$$=2{\int }_0^{\frac{\pi }{2}}{\tan }^5\left(\mathrm{x}\right){\sec }^3\left(\mathrm{x}\right)\mathrm{dx}$$$$=\beta (3,-\frac{7}{2})=\frac{2\mathrm{\Gamma }(-\frac{7}{2})}{\mathrm{\Gamma }(-\frac{1}{2})}$$$$=2(-\frac{2}{7})(-\frac{2}{5})(-\frac{2}{3})$$$$=-\frac{16}{105}\Rightarrow \mid \mathrm{I}\mid =\mid -\frac{16}{105}\mid =\frac{16}{105}$$

Answered by puissant last updated on 05/Jul/21

$$=\mid {\int }_0^{\pi }\frac{{\sin }^5\left(\mathrm{x}\right)}{{\cos }^8\left(\mathrm{x}\right)}\mathrm{dx}\mid =\mid {\int }_0^{\pi }\frac{(1-{2\cos }^2\left(\mathrm{x}\right)+{\cos }^4\left(\mathrm{x}\right))\sin \left(\mathrm{x}\right)}{{\cos }^8\left(\mathrm{x}\right)}\mathrm{dx}\mid$$$$\mathrm{u}=\cos \left(\mathrm{x}\right)\Rightarrow \mathrm{du}=-\sin \left(\mathrm{x}\right)\mathrm{dx}\Rightarrow \mathrm{dx}=-\frac{\mathrm{du}}{\sin \left(\mathrm{x}\right)}$$$$\mid {\int }_1^{-1}\frac{(1-{2\mathrm{u}}^2+{\mathrm{u}}^4)\sin \left(\mathrm{x}\right)}{{\mathrm{u}}^8}\times (-\frac{\mathrm{du}}{\sin \left(\mathrm{x}\right)})\mid$$$$=\mid {\int }_{-1}^1(\frac{1}{{\mathrm{u}}^8}-\frac{2}{{\mathrm{u}}^6}+\frac{1}{{\mathrm{u}}^4})\mathrm{du}\mid =....$$

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