# Problem of the Week

## Updated at Sep 30, 2013 2:11 PM

For this week we've brought you this calculus problem.

How can we solve for the integral of $$\tan^{3}x$$?

Here are the steps:

$\int \tan^{3}x \, dx$

 1 Use Pythagorean Identities: $$\tan^{2}x=\sec^{2}x-1$$.$\int (\sec^{2}x-1)\tan{x} \, dx$2 Expand.$\int \tan{x}\sec^{2}x-\tan{x} \, dx$3 Use Sum Rule: $$\int f(x)+g(x) \, dx=\int f(x) \, dx+\int g(x) \, dx$$.$\int \tan{x}\sec^{2}x \, dx-\int \tan{x} \, dx$4 Use Integration by Substitution on $$\int \tan{x}\sec^{2}x \, dx$$.Let $$u=\tan{x}$$, $$du=\sec^{2}x \, dx$$5 Using $$u$$ and $$du$$ above, rewrite $$\int \tan{x}\sec^{2}x \, dx$$.$\int u \, du$6 Use Power Rule: $$\int {x}^{n} \, dx=\frac{{x}^{n+1}}{n+1}+C$$.$\frac{{u}^{2}}{2}$7 Substitute $$u=\tan{x}$$ back into the original integral.$\frac{\tan^{2}x}{2}$8 Rewrite the integral with the completed substitution.$\frac{\tan^{2}x}{2}-\int \tan{x} \, dx$9 Use Trigonometric Integration: the integral of $$\tan{x}$$ is $$\ln{(\sec{x})}$$.$\frac{\tan^{2}x}{2}-\ln{(\sec{x})}$10 Add constant.$\frac{\tan^{2}x}{2}-\ln{(\sec{x})}+C$Done tan(x)^2/2-ln(sec(x))+C