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