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

1 Answer
Mar 5, 2015

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.