The Notes: `int tan^5(x) sec^3(x) dx` Evaluate the integral

$\displaystyle\int{{\tan}}^{{5}}{\left({x}\right)}{{\sec}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}$

Rewrite the integrand as,

$\displaystyle=\int{{\tan}}^{{4}}{\left({x}\right)}{\tan{{x}}}{\left({x}\right)}{{\sec}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}$

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

$\displaystyle=\int{{\left({{\sec}}^{{2}}{\left({x}\right)}-{1}\right)}}^{{2}}{{\sec}}^{{3}}{\left({x}\right)}{\tan{{\left({x}\right)}}}{\left.{d}{x}\right.}$

Now apply the integral substitution,

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

$\displaystyle{d}{u}={\sec{{\left({x}\right)}}}{\tan{{\left({x}\right)}}}{\left.{d}{x}\right.}$

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

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

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

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

$\displaystyle=\frac{{{u}}^{{7}}}{{7}}-{2}{\left(\frac{{{u}}^{{5}}}{{5}}\right)}+\frac{{{u}}^{{3}}}{{3}}$

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

$\displaystyle=\frac{{1}}{{7}}{{\sec}}^{{7}}{\left({x}\right)}-\frac{{2}}{{5}}{{\sec}}^{{5}}{\left({x}\right)}+\frac{{1}}{{3}}{{\sec}}^{{3}}{\left({x}\right)}$

Add a constant C to the solution,

$\displaystyle=\frac{{1}}{{7}}{{\sec}}^{{7}}{\left({x}\right)}-\frac{{2}}{{5}}{{\sec}}^{{5}}{\left({x}\right)}+\frac{{1}}{{3}}{{\sec}}^{{3}}{\left({x}\right)}+{C}$

The Notes

Wednesday, June 16, 2010

$\displaystyle\int{{\tan}}^{{5}}{\left({x}\right)}{{\sec}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

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