Question Video: Finding the Integration of a Function Involving Trigonometric Functions by Distributing the Division | Nagwa Question Video: Finding the Integration of a Function Involving Trigonometric Functions by Distributing the Division | Nagwa

Question Video: Finding the Integration of a Function Involving Trigonometric Functions by Distributing the Division Mathematics • Second Year of Secondary School

Determine ∫(2 cos³ 2𝑥 + 1/9 cos² 3𝑥) d𝑥.

02:41

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.

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