Video Transcript
Determine the indefinite integral
of two cos cubed three π₯ plus one over nine cos square three π₯ evaluated with
respect to π₯.
This question does, at first
glance, look quite tricky. However, we should spot that we can
actually simplify this quotient somewhat. We essentially reversed the process
we would take when adding two fractions. And we see that we can write the
quotient as two cos cubed three π₯ over nine cos squared three π₯ plus one over nine
cos squared three π₯. The first fraction simplifies to
two-ninths of cos of three π₯. And then to help us spot what to do
next, letβs rewrite the second fraction as a ninth times one over cos squared three
π₯. Next, we recall that the integral
of the sum of two or more functions is actually equal to the sum of the integrals of
those respective functions. And we write this as the integral
of two-ninths of cos of three π₯ evaluated with respect to π₯ plus the integral of a
ninth times one over cos squared three π₯, again, evaluated with respect to π₯.
We also know that we can take any
constant factors outside of the integral and focus on integrating the expression in
π₯. So we write this further as
two-ninths times the integral of cos of three π₯ dπ₯ plus a ninth times the integral
of one over cos squared three π₯ dπ₯. We can quote the general result for
the integral of cos of ππ₯. Itβs one over π sin of ππ₯. And this means the integral of cos
of three π₯ is a third sin of three π₯ plus some constant of integration. Letβs call that π΄. But what do we do about the second
integral? Well, we know the trigonometric
identity one over cos of π₯ equals sec of π₯. And we see that we can rewrite one
over cos squared of three π₯ as sec squared three π₯. And then we have the general result
for the integral of sec squared ππ₯ dπ₯. Itβs one over π tan of ππ₯ plus
some constant. And this means we can write the
integral of sec squared three π₯ as a third tan of three π₯ plus another constant of
integration π΅.
We distribute our parentheses and
we see that two ninths times a third sin of three π₯ is two twenty-sevenths sin of
three π₯. Similarly, we obtain a ninth times
a third of tan three π₯ to be one twenty-seventh of tan of three π₯. And finally, when we multiply each
of our constants by two-ninths and one-ninth, respectively, we end up with a new
constant πΆ. And we find that our integral is
equal to two twenty-sevenths of sin of three π₯ plus one twenty-seventh of tan of
three π₯ plus πΆ.
