How do you find the integral of #cotx^5cscx^2#?

Question

Cameron Alexander · Accepted Answer

Our required integral is,

#I = intcot^5xcsc^2xdx#

First, we make the substitution,

#cotx=t#

If we differentiate both sides with respect to #t#,

#d/dt(cotx)=dt/dt#

Then we get,

#(-csc^2x)dx/dt=1#

On rearraging,

#csc^2xdx = -dt#

On substituting these values in our original integral, we get

#I = intt^5dt#

which can be solved trivially to get the solution

#t^6/6 + C#

On substituting the value of #t = cotx#, we get,

#I = cot^6x/6+C#

In case your question was #intcotx^5cscx^2dx# the answer is far more complicated and beyond the scope of a straight-forward pen and paper solution. Maybe, the solution does not even exist.

William Abdalla · Answer

undefined To integrate $ \cot(x)^5 \csc(x)^2 $, you can use the substitution method. Let $ u = \cot(x) $. Then, $ du = -\csc(x)^2 dx $.

So, $ dx = -\frac{du}{\csc(x)^2} = -du \sin(x) $, and $ \cot(x) = u $.

Substituting these into the integral:

$ \int u^5 (-du \sin(x)) $

$ = -\int u^5 du \sin(x) $

Now, integrate $ u^5 $ with respect to $ u $:

$ = -\frac{1}{6}u^6 \sin(x) + C $

Replace $ u $ with $ \cot(x) $:

$ = -\frac{1}{6}\cot(x)^6 \sin(x) + C $

So, the integral of $ \cot(x)^5 \csc(x)^2 $ is $ -\frac{1}{6}\cot(x)^6 \sin(x) + C $.