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Question Number 167666 by LEKOUMA last updated on 22/Mar/22

$$I_n=\int \frac{dx}{\cos {}^{n}x}$$$$Provethat$$$$I_n=\frac{n-2}{n-1}{I}_{n-2}+\frac{\sin x}{(n-1)\cos {}^{n-1}x}$$

Commented by peter frank last updated on 22/Mar/22

$$\mathrm{Reduction}\mathrm{formular}$$

Answered by chhaythean last updated on 22/Mar/22

$$\mathrm{Solution}$$$${\mathrm{I}}_{\mathrm{n}}=\int \frac{\mathrm{dx}}{{\cos }^{\mathrm{n}}\mathrm{x}}=\int {\sec }^{\mathrm{n}}\mathrm{xdx}$$$$=\int {\sec }^{\mathrm{n}-2}{\mathrm{xsec}}^2\mathrm{xdx}$$$$\mathrm{let}\{\begin{array}{l}\mathrm{u}={\sec }^{\mathrm{n}-2}\mathrm{x}\Rightarrow \mathrm{du}=(\mathrm{n}-2){\sec }^{\mathrm{n}-2}\mathrm{xtanxdx}\\ \mathrm{dv}={\sec }^2\mathrm{xdx}\Rightarrow \mathrm{v}=\mathrm{tanx}\end{array}$$$${\mathrm{I}}_{\mathrm{n}}={\sec }^{\mathrm{n}-2}\mathrm{xtanx}-(\mathrm{n}-2)\int {\sec }^{\mathrm{n}-2}{\mathrm{xtan}}^2\mathrm{xdx}$$$$={\sec }^{\mathrm{n}-2}\mathrm{xtanx}-(\mathrm{n}-2)\int {\sec }^{\mathrm{n}}\mathrm{xdx}+(\mathrm{n}-2)\int {\sec }^{\mathrm{n}-2}\mathrm{dx}$$$${\mathrm{I}}_{\mathrm{n}}={\sec }^{\mathrm{n}-2}\mathrm{xtanx}-(\mathrm{n}-2){\mathrm{I}}_{\mathrm{n}}+(\mathrm{n}-2){\mathrm{I}}_{\mathrm{n}-2}$$$${\mathrm{I}}_{\mathrm{n}}+(\mathrm{n}-2){\mathrm{I}}_{\mathrm{n}}={\sec }^{\mathrm{n}-2}\mathrm{xtanx}+(\mathrm{n}-2){\mathrm{I}}_{\mathrm{n}-2}$$$$(\mathrm{n}-1){\mathrm{I}}_{\mathrm{n}}={\sec }^{\mathrm{n}-2}\mathrm{xtanx}+(\mathrm{n}-2){\mathrm{I}}_{\mathrm{n}-2}$$$${\mathrm{I}}_{\mathrm{n}}=\frac{\frac{1}{{\cos }^{\mathrm{n}-2}\mathrm{x}}\times \frac{\mathrm{sinx}}{\mathrm{cosx}}}{\mathrm{n}-1}+\frac{\mathrm{n}-2}{\mathrm{n}-1}{\mathrm{I}}_{\mathrm{n}-2}$$$${\mathrm{I}}_{\mathrm{n}}=\frac{\mathrm{n}-2}{\mathrm{n}-1}{\mathrm{I}}_{\mathrm{n}-2}+\frac{\mathrm{sinx}}{{\cos }^{\mathrm{n}-1}\mathrm{x}(\mathrm{n}-1)}\mathrm{true}$$$$\mathrm{So}\overline{)\begin{array}{c}{\mathrm{I}}_{\mathrm{n}}=\frac{\mathrm{n}-2}{\mathrm{n}-1}{\mathrm{I}}_{\mathrm{n}-2}+\frac{\mathrm{sinx}}{(\mathrm{n}-1){\cos }^{\mathrm{n}-1}\mathrm{x}}\mathrm{is}\mathrm{proved}.\end{array}}$$

Commented by LEKOUMA last updated on 22/Mar/22

$$Thanks$$

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