# Problem of the Week

## Updated at Oct 7, 2013 8:24 AM

To get more practice in calculus, we brought you this problem of the week:

How can we find the integral of $$\csc^{3}x$$?

Check out the solution below!

$\int \csc^{3}x \, dx$

 1 Use Integration by Parts on $$\int \csc^{3}x \, dx$$.Let $$u=\csc{x}$$, $$dv=\csc^{2}x$$, $$du=-\csc{x}\cot{x} \, dx$$, $$v=-\cot{x}$$2 Substitute the above into $$uv-\int v \, du$$.$-\csc{x}\cot{x}-\int \cot^{2}x\csc{x} \, dx$3 Use Pythagorean Identities: $$\cot^{2}x=\csc^{2}x-1$$.$-\csc{x}\cot{x}-\int (\csc^{2}x-1)\csc{x} \, dx$4 Expand.$-\csc{x}\cot{x}-\int \csc^{3}x-\csc{x} \, dx$5 Use Sum Rule: $$\int f(x)+g(x) \, dx=\int f(x) \, dx+\int g(x) \, dx$$.$-\csc{x}\cot{x}-\int \csc^{3}x \, dx+\int \csc{x} \, dx$6 Set it as equal to the original integral $$\int \csc^{3}x \, dx$$.$\int \csc^{3}x \, dx=-\csc{x}\cot{x}-\int \csc^{3}x \, dx+\int \csc{x} \, dx$7 Add $$\int \csc^{3}x \, dx$$ to both sides.$\int \csc^{3}x \, dx+\int \csc^{3}x \, dx=-\csc{x}\cot{x}+\int \csc{x} \, dx$8 Simplify  $$\int \csc^{3}x \, dx+\int \csc^{3}x \, dx$$  to  $$2\int \csc^{3}x \, dx$$.$2\int \csc^{3}x \, dx=-\csc{x}\cot{x}+\int \csc{x} \, dx$9 Divide both sides by $$2$$.$\int \csc^{3}x \, dx=\frac{-\csc{x}\cot{x}+\int \csc{x} \, dx}{2}$10 Original integral solved.$\frac{-\csc{x}\cot{x}+\int \csc{x} \, dx}{2}$11 Use Trigonometric Integration: the integral of $$\csc{x}$$ is $$\ln{(\csc{x}-\cot{x})}$$.$\frac{-\csc{x}\cot{x}+\ln{(\csc{x}-\cot{x})}}{2}$12 Add constant.$\frac{-\csc{x}\cot{x}+\ln{(\csc{x}-\cot{x})}}{2}+C$Done(-csc(x)*cot(x)+ln(csc(x)-cot(x)))/2+C