Problem of the Week

Updated at Feb 10, 2014 3:33 PM

This week we have another calculus problem:

How can we find the integral of \(\cot^{3}x\)?

Let's start!



\[\int \cot^{3}x \, dx\]

1
Use Pythagorean Identities: \(\cot^{2}x=\csc^{2}x-1\).
\[\int (\csc^{2}x-1)\cot{x} \, dx\]

2
Expand.
\[\int \cot{x}\csc^{2}x-\cot{x} \, dx\]

3
Use Sum Rule: \(\int f(x)+g(x) \, dx=\int f(x) \, dx+\int g(x) \, dx\).
\[\int \cot{x}\csc^{2}x \, dx-\int \cot{x} \, dx\]

4
Simplify the trigonometric functions.
\[\int \frac{\cos{x}}{\sin^{3}x} \, dx-\int \cot{x} \, dx\]

5
Use Integration by Substitution on \(\int \frac{\cos{x}}{\sin^{3}x} \, dx\).
Let \(u=\sin{x}\), \(du=\cos{x} \, dx\)

6
Using \(u\) and \(du\) above, rewrite \(\int \frac{\cos{x}}{\sin^{3}x} \, dx\).
\[\int \frac{1}{{u}^{3}} \, du\]

7
Use Power Rule: \(\int {x}^{n} \, dx=\frac{{x}^{n+1}}{n+1}+C\).
\[-\frac{1}{2{u}^{2}}\]

8
Substitute \(u=\sin{x}\) back into the original integral.
\[-\frac{1}{2\sin^{2}x}\]

9
Rewrite the integral with the completed substitution.
\[-\frac{1}{2\sin^{2}x}-\int \cot{x} \, dx\]

10
Use Trigonometric Integration: the integral of \(\cot{x}\) is \(\ln{(\sin{x})}\).
\[-\frac{1}{2\sin^{2}x}-\ln{(\sin{x})}\]

11
Add constant.
\[-\frac{1}{2\sin^{2}x}-\ln{(\sin{x})}+C\]

Done
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