How do you find #int cot^2x*secxdx#?

1 Answer
Dec 1, 2015

Notice that #cot^2(x)sec(x) = cot(x)csc(x)# and perform a simple substitution to find that
#intcot^2(x)sec(x)dx = -csc(x) + C#

Explanation:

When an easy substitution does not immediately jump out when looking at an integral, often one can manipulate the integrand into a more convenient form.

In this case:

#intcot^2(x)sec(x)dx = int(cos^2(x))/(sin^2(x))1/cos(x)dx#
#= intcos(x)/sin(x)1/sin(x)dx#
#= intcot(x)csc(x)dx#

We could go directly to the answer from here, as we know that #csc(x)# is the antiderivative of #-csc(x)cot(x)# but let's do the substitution just to see how it works out.

Let #u = csc(x) => du = -csc(x)cot(x)dx#

#=> intcot(x)csc(x)dx = int-du#
# = -intdu#
#= -u+C#
#= -csc(x) + C#

Thus #intcot^2(x)sec(x)dx = -csc(x)+C#