Problem of the Week

Updated at Sep 23, 2013 4:12 PM

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

Below is the solution.



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

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

2
Use Integration by Substitution.
Let \(u=\cos{x}\), \(du=-\sin{x} \, dx\)

3
Using \(u\) and \(du\) above, rewrite \(\int (1-\cos^{2}x)\sin{x} \, dx\).
\[\int -(1-{u}^{2}) \, du\]

4
Use Power Rule: \(\int {x}^{n} \, dx=\frac{{x}^{n+1}}{n+1}+C\).
\[\frac{{u}^{3}}{3}-u\]

5
Substitute \(u=\cos{x}\) back into the original integral.
\[\frac{\cos^{3}x}{3}-\cos{x}\]

6
Add constant.
\[\frac{\cos^{3}x}{3}-\cos{x}+C\]

Done
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