How do you integrate # csc^3x#?

Question

Elijah Abram · Accepted Answer

#(-cotxcscx-ln(abs(cotx+cscx)))/2+C#

We have:

#I=intcsc^3xdx#

First, rewrite the integral as follows in order to apply integration by parts:

#I=intcsc^2xcscxdx#

Since integration by parts takes the form #intudv=uv-intvdu#, let:

#{(u=cscx" "=>" "du=-cotxcscxdx),(dv=csc^2xdx" "=>" "v=-cotx):}#

Utilizing integration by components

#I=-cotxcscx-intcot^2xcscxdx#

Through the Pythagorean identity, write #cot^2x# as #csc^2x-1#.

#I=-cotxcscx-int(csc^2x-1)(cscx)dx#

#I=-cotxcscx-intcsc^3xdx+intcscxdx#

Note that #I=intcsc^3xdx# and #intcscxdx=-ln(abs(cotx+cscx))#.

#I=-cotxcscx-I-ln(abs(cotx+cscx))#

Add the original integral #I# to both sides.

#2I=-cotxcscx-ln(abs(cotx+cscx))#

Solve for #I# and add the constant of integration:

#I=(-cotxcscx-ln(abs(cotx+cscx)))/2+C#

Elijah Abram · Answer

undefined To integrate $ \csc^3(x) $, you can use trigonometric substitution. Let $ u = \csc(x) $ and $ du = -\csc(x) \cot(x) \, dx $. Then, $ \int \csc^3(x) \, dx $ becomes $ \int -u^2 \, du $, which can be easily integrated to get $ -\frac{1}{2}u^3 + C $. Finally, substitute $ u = \csc(x) $ back in to get the final result: $ -\frac{1}{2}\csc^3(x) + C $.