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

Wednesday, January 30, 2013

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

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

To solve, apply the Pythagorean identity $\displaystyle{{\tan}}^{{2}}{\left({x}\right)}={{\sec}}^{{2}}{\left({x}\right)}-{1}$ until the integrand is in the form $\displaystyle\int{{u}}^{{n}}{d}{u}$ .

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

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

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

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

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

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

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

For the first and second integral, apply u-substitution method. Let u be:

$\displaystyle{u}={\tan{{x}}}$

Then, differentiate u.

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

Plugging them, the first and second integral becomes:

$\displaystyle=\int{{u}}^{{3}}{d}{u}-\int{u}{d}{u}+\int{\tan{{\left({x}\right)}}}{\left.{d}{x}\right.}$

Then, apply the integral formula $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$ and $\displaystyle\int{\tan{{\left(\theta\right)}}}{d}\theta={\ln}{\left|{\sec{{\left(\theta\right)}}}\right|}+{C}$ .

$\displaystyle=\frac{{{u}}^{{4}}}{{4}}-\frac{{{u}}^{{2}}}{{2}}+{\ln}{\left|{\sec{{\left({x}\right)}}}\right|}+{C}$

And, substitute back u = tan(x).

$\displaystyle=\frac{{{{\tan}}^{{4}}{\left({x}\right)}}}{{4}}-\frac{{{{\tan}}^{{2}}{\left({x}\right)}}}{{2}}+{\ln}{\left|{\sec{{\left({x}\right)}}}\right|}+{C}$

Therefore, $\displaystyle\int{{\tan}}^{{5}}{\left({x}\right)}{\left.{d}{x}\right.}=\frac{{{{\tan}}^{{4}}{\left({x}\right)}}}{{4}}-\frac{{{{\tan}}^{{2}}{\left({x}\right)}}}{{2}}+{\ln}{\left|{\sec{{\left({x}\right)}}}\right|}+{C}$ .

The Notes

Wednesday, January 30, 2013

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

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