Video: Integrating Reciprocal Trigonometric Functions

Determine ∫ (βˆ’5/(4 cosΒ² 9π‘₯)) dπ‘₯.

02:40

Video Transcript

Determine the indefinite integral of negative five over four cos squared nine π‘₯ with respect to π‘₯.

Let’s have a look at the function we’ve been asked to integrate. It includes a trigonometric function cos of nine π‘₯ which has been squared. And, in fact, we have a reciprocal trigonometric function because this factor of cos squared nine π‘₯ is in the denominator of a quotient. First then, we recall one of the definitions of reciprocal trigonometric functions, which is that one over cos π‘₯ is equal to sec π‘₯. And, in fact, if we were to square each of these, we have that one over cos squared π‘₯ is equal to sec squared π‘₯.

Changing the argument to nine π‘₯ just means we have sec squared nine π‘₯ rather than sec squared π‘₯. And we can bring the factor of negative five over four out the front of our integral as a multiplicative constant. So this integral is equal to negative five over four, the integral of sec squared nine π‘₯ with respect to π‘₯. Now, in order to perform this integral, we need to recall one of our standard derivatives, which is that the derivative with respect to π‘₯ of tan π‘₯ is equal to sec squared π‘₯. And this can be proved by writing tan π‘₯ as sin π‘₯ over cos π‘₯ and applying the quotient rule.

If, instead of tan π‘₯, we were differentiating tan of π‘Žπ‘₯ for some constant π‘Ž, then this is equal to a sec squared π‘Žπ‘₯. We can divide both sides of this by π‘Ž and then bring the factor of one over π‘Ž on the left inside the derivative. And we see that the derivative with respect to π‘₯ of one over π‘Ž tan π‘Žπ‘₯ is equal to sec squared π‘Žπ‘₯. Or, we see that the antiderivative of sec squared of π‘Žπ‘₯ is equal to one over π‘Ž tan π‘Žπ‘₯.

By applying this result then, we can say that the integral of sec squared of nine π‘₯ with respect to π‘₯ is equal to one-ninth of tan nine π‘₯. And we must remember to include a constant of integration 𝐢 as we were performing an indefinite integral. We can simplify the constant. And we have that this integral is equal to negative five over 36 tan nine π‘₯ plus a constant of integration 𝐢.

These general results for integrating and differentiating trigonometric functions are important. And we need to remember them. Written in another way, we could say that the integral of sec squared of π‘Žπ‘₯ with respect to π‘₯ is equal to one over π‘Ž tan π‘Žπ‘₯ plus 𝐢. And then applying this general result is what got us to our answer.

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