Determine the indefinite integral
of nine tan three 𝑥 sec three 𝑥 with respect to 𝑥.
The integrand here contains the
product of a tangent and a secant, both of which have the same argument three
𝑥. We can recall the following
standard integral. The indefinite integral of sec 𝑥
tan 𝑥 with respect to 𝑥 is equal to sec of 𝑥 plus 𝐶. To use this result in this example,
we need to modify the argument three 𝑥 of the trigonometric functions. We can do this by making a
substitution. We’ll let 𝑢 equal three 𝑥.
It follows then that d𝑢 by d𝑥 is
equal to three. And whilst d𝑢 by d𝑥 is not a
fraction, we can treat it a little like one. So equivalently d𝑢 is equal to
three d𝑥, or one-third d𝑢 is equal to d𝑥. We can now perform this
substitution replacing three 𝑥 with 𝑢 and d𝑥 with one-third d𝑢 to obtain the
indefinite integral of nine tan 𝑢 sec 𝑢 one-third d𝑢. The constant in the integrand
simplifies to three, and we can then bring this factor of three out the front to
give three multiplied by the indefinite integral of tan 𝑢 sec 𝑢 with respect to
By the standard result, this
integrates to three sec 𝑢 plus a constant of integration 𝐶. All that remains is to undo our
substitution, so we need to replace 𝑢 with three 𝑥, which gives our final answer
to the problem. We’ve found that the indefinite
integral of nine tan three 𝑥 sec three 𝑥 with respect to 𝑥 is equal to three sec
three 𝑥 plus a constant of integration 𝐶.