How do you find the integral of #int cot^4 (3x)dx#?

Question

How do you find the integral of #int  cot^4 (3x)dx#?

Ezra Adams · Accepted Answer

#intcot^4(3x)dx=1/9[9x+3cot(3x)-cot^3(3x)]+c#

Here ,

#I=intcot^4(3x)dx#

Subst. #color(blue)(3x=u=>3dx=du=>dx=1/3du#

#I=1/3intcot^4udu#

#I=1/3intcot^2u*cot^2udu#

#I=1/3intcot^2u(csc^2u-1)duto[becausecolor(red)(csc^2theta-cot^2theta=1)]#

#I=1/3intcot^2ucsc^2udu-1/3intcot^2udu#

#I=-1/3intcot^2u(-csc^2u)du-1/3intcot^2udu#

#I=-1/3int(cotu)^2d(cotu)-1/3int(csc^2u-1)du#

#I=-1/3(cotu)^3/3-1/3{-cotu-u}+c#

#I=-1/9cot^3u+1/3cotu+1/3u+c#

#I=1/9[3u+3cotu-cot^3u]+c#

Subst. back #color(blue)(u=3x#

#I=1/9[9x+3cot(3x)-cot^3(3x)]+c#

Charles Anctil · Answer

undefined To find the integral of $\int \cot^4(3x) \, dx$, you can use the following steps:

1. Rewrite $\cot^4(3x)$ as $(\cot^2(3x))^2$.
2. Use the Pythagorean identity for cotangent: $\cot^2(3x) = \csc^2(3x) - 1$.
3. Substitute the identity into the integral to get $(\csc^2(3x) - 1)^2$.
4. Expand and simplify the expression to get the integrand in terms of $\csc(3x)$ and its derivatives.
5. Perform the integral term by term.
6. Finally, substitute back $\cot(3x)$ for $\csc(3x)$ using the reciprocal identity for cotangent.

The process involves multiple steps of substitution, expansion, and integration by parts.