The Notes: `int tan^4(x) sec^6(x) dx` Evaluate the integral

$\displaystyle\int{{\tan}}^{{4}}{\left({x}\right)}{{\sec}}^{{6}}{\left({x}\right)}{\left.{d}{x}\right.}$

Rewrite the integrand as,

$\displaystyle=\int{{\tan}}^{{4}}{\left({x}\right)}{{\sec}}^{{2}}{\left({x}\right)}{{\sec}}^{{4}}{\left({x}\right)}{\left.{d}{x}\right.}$

use the identity: $\displaystyle{{\sec}}^{{2}}{\left({x}\right)}={1}+{{\tan}}^{{2}}{\left({x}\right)}$

$\displaystyle=\int{{\tan}}^{{4}}{\left({x}\right)}{{\sec}}^{{2}}{\left({x}\right)}{{\left({1}+{{\tan}}^{{2}}{\left({x}\right)}\right)}}^{{2}}{\left.{d}{x}\right.}$

Now apply integral substitution,

Let $\displaystyle{u}={\tan{{\left({x}\right)}}}$

$\displaystyle\Rightarrow{d}{u}={{\sec}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\int{{u}}^{{4}}{{\left({1}+{{u}}^{{2}}\right)}}^{{2}}{d}{u}$

$\displaystyle=\int{{u}}^{{4}}{\left({1}+{2}{{u}}^{{2}}+{{u}}^{{4}}\right)}{d}{u}$

$\displaystyle=\int{\left({{u}}^{{4}}+{2}{{u}}^{{6}}+{{u}}^{{8}}\right)}{d}{u}$

$\displaystyle=\int{{u}}^{{4}}{d}{u}+{2}\int{{u}}^{{6}}{d}{u}+\int{{u}}^{{8}}{d}{u}$

$\displaystyle=\frac{{{u}}^{{5}}}{{5}}+{2}{\left(\frac{{{u}}^{{7}}}{{7}}\right)}+\frac{{{u}}^{{9}}}{{9}}$

Substitute back $\displaystyle{u}={\tan{{\left({x}\right)}}}$

$\displaystyle=\frac{{1}}{{5}}{{\tan}}^{{5}}{\left({x}\right)}+\frac{{2}}{{7}}{{\tan}}^{{7}}{\left({x}\right)}+\frac{{1}}{{9}}{{\tan}}^{{9}}{\left({x}\right)}$

Add a constant C to the solution,

$\displaystyle=\frac{{1}}{{5}}{{\tan}}^{{5}}{\left({x}\right)}+\frac{{2}}{{7}}{{\tan}}^{{7}}{\left({x}\right)}+\frac{{1}}{{9}}{{\tan}}^{{9}}{\left({x}\right)}+{C}$

The Notes

Thursday, October 11, 2012

$\displaystyle\int{{\tan}}^{{4}}{\left({x}\right)}{{\sec}}^{{6}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

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