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

Determine ∫ (2 cosΒ³ 3π‘₯ + 1)/(9 cosΒ² 3π‘₯) dπ‘₯.

02:33

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, let’s 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 π‘Žπ‘₯. It’s one over π‘Ž sin of π‘Žπ‘₯. And this means the integral of cos of three π‘₯ is a third sin of three π‘₯ plus some constant of integration. Let’s call that 𝐴. 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 π‘Žπ‘₯ dπ‘₯. It’s one over π‘Ž tan of π‘Žπ‘₯ 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 𝐡.

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 𝐢. 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 𝐢.

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