The Notes: `int (tan^2(x) + tan^4(x)) dx` Evaluate the integral

Wednesday, October 12, 2016

$\displaystyle\int{\left({{\tan}}^{{2}}{\left({x}\right)}+{{\tan}}^{{4}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

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

Factor out the GCF of the two terms.

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

Then, apply the trigonometric identity $\displaystyle{{\sec}}^{{2}}\theta={{\tan}}^{{2}}\theta+{1}$ .

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

To take the integral of this, use u-substitution method. So let u be:

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

Then, differentiate u.

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

Plugging them, the integral becomes:

$\displaystyle=\int{{u}}^{{2}}{d}{u}$

Then, apply the formula $\displaystyle\int{{u}}^{{n}}{d}{u}=\frac{{{u}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$ .

$\displaystyle=\frac{{{u}}^{{3}}}{{3}}+{C}$

And, substitute back u = tan(x).

$\displaystyle=\frac{{{{\tan}}^{{3}}{\left({x}\right)}}}{{3}}+{C}$

Therefore, $\displaystyle\int{\left({{\tan}}^{{2}}{\left({x}\right)}+{{\tan}}^{{4}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}=\frac{{{{\tan}}^{{3}}{\left({x}\right)}}}{{3}}+{C}$ .

The Notes

Wednesday, October 12, 2016

$\displaystyle\int{\left({{\tan}}^{{2}}{\left({x}\right)}+{{\tan}}^{{4}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}$ Evaluate the integral

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