Video Transcript
Determine the indefinite integral
of two cos cubed of three 𝑥 plus one over nine cos squared of three 𝑥 with respect
to 𝑥.
Now at first glance, this integral
looks quite complicated. We should spot, however, that we
can actually simplify the quotient. How we do this is to essentially
reverse the process we follow when adding two fractions. And we see that we can write the
quotient as two cos cubed three 𝑥 over nine cos squared three 𝑥 plus one over nine
cos squared three 𝑥. By dividing through in the first
fraction by cos squared three 𝑥, our first term simplifies to two over nine cos
three 𝑥. And then to help us to spot what to
do next, let’s rewrite the second fraction as a ninth times one over cos squared of
three 𝑥.
Next, we recall that the integral
of the sum of two or more functions is the sum of their integrals. And we also know that we can take
any constant factors outside of the integral and focus on integrating the expression
in 𝑥. And applying these properties gives
us two over nine times the integral of the cos of three 𝑥 with respect to 𝑥 plus
one over nine times the integral of one over the cos squared of three 𝑥 again with
respect to 𝑥. We can quote the result for the
integral of cos of 𝑎𝑥. It’s one over 𝑎 times the sin of
𝑎𝑥 plus the constant of integration 𝑐. And this means that the integral of
cos of three 𝑥 is a third sin of three 𝑥 plus some constant of integration which
we’ll call 𝐴.
But what do we do about the second
integral? Well, we know the trigonometric
identity one over cos of 𝑥 is equal to sec 𝑥. And so we see that we can rewrite
one over cos squared of three 𝑥 as sec squared of three 𝑥. So now we can use the general
result for the integral of sec squared of 𝑎𝑥 with respect to 𝑥. And that’s one over 𝑎 times the
tan of 𝑎𝑥 plus some constant 𝑐. And this means we can write the
integral of sec squared three 𝑥 with respect to 𝑥 as one over three tan three 𝑥
plus a constant of integration which we’ll call 𝐵.
Finally, we distribute our
parentheses and we see that two over nine times one-third of sin three 𝑥 is equal
to two over 27 times sin three 𝑥. And one over nine times one over
three tan three 𝑥 is equal to one over 27 tan three 𝑥. We multiply each of our constants
by two over nine and one over nine, respectively. And we end up with a new constant
𝐶. We find then that our integral is
two over 27 times the sin of three 𝑥 plus one over 27 times tan of three 𝑥 plus
the constant of integration 𝐶.
