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
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