Integral of sec^n(x)dx

Sunday, February 2, 2014

Integral of sec^n(x)dx

You need to use reduction formula to integrate the function, such that:

$\displaystyle\int{{\sec}}^{{n}}{x}{\left.{d}{x}\right.}=\int{{\sec}}^{{{n}-{2}}}{x}\cdot{{\sec}}^{{2}}{x}{\left.{d}{x}\right.}$

You need to use integration by parts, such that:

$\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$

$\displaystyle{u}={{\sec}}^{{{n}-{2}}}{x}\Rightarrow{d}{u}={\left({n}-{2}\right)}\cdot{{\sec}}^{{{n}-{3}}}{x}{\tan{{\left({x}\right)}}}$

$\displaystyle{d}{v}={{\sec}}^{{2}}{x}\Rightarrow{v}={\tan{{x}}}$

$\displaystyle\int{{\sec}}^{{{n}-{2}}}{x}\cdot{{\sec}}^{{2}}{x}{\left.{d}{x}\right.}={\tan{{x}}}\cdot{{\sec}}^{{{n}-{2}}}{x}-{\left({n}-{2}\right)}\cdot\int{{\sec}}^{{{n}-{3}}}{x}\cdot{\tan{{x}}}{\left.{d}{x}\right.}$

You need to replace $\displaystyle{{\sec}}^{{2}}{x}-{1}$ for $\displaystyle{{\tan}}^{{2}}{x}$ , such that:

$\displaystyle\int{{\sec}}^{{{n}-{2}}}{x}\cdot{{\sec}}^{{2}}{x}{\left.{d}{x}\right.}={\tan{{x}}}\cdot{{\sec}}^{{{n}-{2}}}{x}-{\left({n}-{2}\right)}\cdot\int{\left({{\sec}}^{{n}}{x}-{{\sec}}^{{{n}-{2}}}{x}\right)}{\left.{d}{x}\right.}$

You need to put $\displaystyle{I}_{{n}}=\int{{\sec}}^{{n}}{x}{\left.{d}{x}\right.},$ such that:

$\displaystyle{I}_{{n}}={\tan{{x}}}\cdot{{\sec}}^{{{n}-{2}}}{x}-{\left({n}-{2}\right)}\cdot{I}_{{n}}+{\left({n}-{2}\right)}\cdot\int{{\sec}}^{{{n}-{2}}}{x}{\left.{d}{x}\right.}$

$\displaystyle{I}_{{n}}+{\left({n}-{2}\right)}\cdot{I}_{{n}}={\tan{{x}}}\cdot{{\sec}}^{{{n}-{2}}}{x}+{\left({n}-{2}\right)}\cdot\int{{\sec}}^{{{n}-{2}}}{x}{\left.{d}{x}\right.}$

$\displaystyle{I}_{{n}}\cdot{\left({n}-{2}+{1}\right)}={\tan{{x}}}\cdot{{\sec}}^{{{n}-{2}}}{x}+{\left({n}-{2}\right)}\cdot\int{{\sec}}^{{{n}-{2}}}{x}{\left.{d}{x}\right.}$

Put $\displaystyle\int{{\sec}}^{{{n}-{2}}}{x}{\left.{d}{x}\right.}={I}_{{{n}-{2}}}$

$\displaystyle{\left({n}-{1}\right)}\cdot{I}_{{n}}={\tan{{x}}}\cdot{{\sec}}^{{{n}-{2}}}{x}+{\left({n}-{2}\right)}\cdot{I}_{{{n}-{2}}}$

$\displaystyle{I}_{{n}}=\frac{{1}}{{{n}-{1}}}\cdot{\tan{{x}}}\cdot{{\sec}}^{{{n}-{2}}}{x}+\frac{{{n}-{2}}}{{{n}-{1}}}\cdot{I}_{{{n}-{2}}}$

Hence, evaluating the given integral, using reduction formula yields $\displaystyle{I}_{{n}}=\frac{{1}}{{{n}-{1}}}\cdot{\tan{{x}}}\cdot{{\sec}}^{{{n}-{2}}}{x}+\frac{{{n}-{2}}}{{{n}-{1}}}\cdot{I}_{{{n}-{2}}}.$

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Sunday, February 2, 2014