### Video Transcript

Determine the indefinite integral of negative seven sec squared three 𝑥 over five
tan three 𝑥 with respect to 𝑥.

In this question, we are asked to integrate a quotient of trigonometric
functions. Recall that the derivative of tan 𝑥 with respect to 𝑥 is sec squared 𝑥. Using the chain rule, we can deduce that for all constants 𝑎, the derivative of tan
𝑎𝑥 with respect to 𝑥 is 𝑎 sec squared 𝑎𝑥. Therefore, the derivative of tan three 𝑥 with respect to 𝑥 is three sec squared
three 𝑥. Using this, we note that the derivative of the function tan three 𝑥 in the
denominator of the integrand of the integral we are asked to evaluate almost appears
in the numerator. The difference is that instead of three sec squared three 𝑥, we have negative seven
over five sec squared three 𝑥 in the numerator.

In order to evaluate this integral, we will use the following property. Given a nonzero function 𝑓 for all constants 𝑎, the indefinite integral of 𝑎
multiplied by the derivative of 𝑓 with respect to 𝑥 𝑓 prime of 𝑥 over 𝑓 of 𝑥
with respect to 𝑥 is equal to 𝑎 multiplied by the natural logarithm of the
absolute value of 𝑓 of 𝑥 plus 𝐶, where 𝐶 is the constant of integration. In order to use this result, let 𝑓 of 𝑥 equal tan of three 𝑥. Then, the derivative of 𝑓 with respect to 𝑥, 𝑓 prime of 𝑥, is equal to three sec
squared three 𝑥, as we have seen earlier.

We’re going to try to manipulate this equation in order to write the numerator of our
integrand in the form 𝑎 multiplied by 𝑓 prime of 𝑥 for some constant 𝑎. Dividing both sides of the equation by three, we obtain that one over three
multiplied by 𝑓 prime of 𝑥 equals sec squared three 𝑥. Multiplying both sides of the resulting equation by negative seven over five, we
obtain negative seven over 15 multiplied by 𝑓 prime of 𝑥 equals negative seven
over five multiplied by sec squared of three 𝑥. Therefore, we can rewrite the integral we’re asked to determine as the indefinite
integral of negative seven over 15 multiplied by 𝑓 prime of 𝑥 over 𝑓 of 𝑥 with
respect to 𝑥, where 𝑓 of 𝑥 equals tan of three 𝑥.

Letting 𝑎 equal negative seven over 15 in the result described earlier, we obtain
that the integral in question evaluates to negative seven over 15 multiplied by the
natural logarithm of the absolute value of 𝑓 of 𝑥 plus 𝐶, where 𝐶 is the
constant of integration. Substituting 𝑓 of 𝑥 equals tan of three 𝑥, we obtain that the integral in question
is equal to negative seven over 15 multiplied by the natural logarithm of the
absolute value of the function tan of three 𝑥 plus 𝐶, where 𝐶 is the constant of
integration. This is our final answer.