# 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$Donecos(x)^3/3-cos(x)+C