# 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 -1/(2*sin(x)^2)-ln(sin(x))+C