Question Video: Integrating Trigonometric Functions Involving Reciprocal Trigonometric Functions Mathematics • Higher Education

Determine ∫ (βˆ’4 cos π‘₯ + (6/cosΒ² π‘₯) βˆ’ 20) dπ‘₯.

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Video Transcript

Determine the integral of negative four cos of π‘₯ plus six divided by the cos squared of π‘₯ minus 20 with respect to π‘₯.

In this question, we’re asked to evaluate the integral of a function consisting of three terms. The first term is a trigonometric function, and the third term is a constant term. We know how to integrate both of these functions. The second term is a trigonometric function; however, it’s a reciprocal trigonometric function because cosine appears in the denominator of our function. So we’ll start by rewriting this by using the reciprocal trigonometric identities.

We recall that one divided by the cos of π‘₯ is defined to be equal to the sec of π‘₯. If we then square both sides of this equation, we see that one divided by cos squared of π‘₯ is equal to sec squared of π‘₯. And we have six divided by the cos squared of π‘₯, so this is equal to six sec squared of π‘₯. Therefore, we can rewrite our integral as the integral of negative four cos of π‘₯ plus six sec squared of π‘₯ minus 20 with respect to π‘₯.

And now we can evaluate the integral of each of these terms separately. Let’s start with the first term, the integral of negative four cos of π‘₯. To do this, we recall the following result. For any real constant π‘Ž, the integral of π‘Ž cos of π‘₯ with respect to π‘₯ is π‘Ž sin of π‘₯ plus a constant of integration 𝐢. Our value of π‘Ž is negative four, so we substitute this into our formula to get negative four sin of π‘₯. And we could add a constant of integration. However, we can just add one at the end of our expression.

We can integrate our second term by using another result. We recall that for any real constant π‘Ž, the integral of π‘Ž sec squared of π‘₯ with respect to π‘₯ is π‘Ž tan of π‘₯ plus a constant of integration 𝐢. This time, our coefficient value of π‘Ž is six. So we substitute this value into our formula to get six tan of π‘₯. Finally, we want to evaluate the integral of negative 20 with respect to π‘₯. There’s a few different ways of doing this. The easiest way is to remember that when we evaluate the integral of a constant, we just need to multiply it by our variable. The integral of negative 20 with respect to π‘₯ is negative 20π‘₯ plus a constant of integration 𝐢.

And this is because the slope of a linear function is just the coefficient of π‘₯. So the slope of negative 20π‘₯ is negative 20. It’s an antiderivative of negative 20. This then gives us negative four sin π‘₯ plus six tan of π‘₯ minus 20π‘₯ plus a constant of integration 𝐢. And finally, we can rewrite this by writing the negative 20π‘₯ term at the beginning of our expression. And this gives us our final result. The integral of negative four cos of π‘₯ plus six divided by cos squared of π‘₯ minus 20 with respect to π‘₯ is equal to negative 20π‘₯ minus four sin π‘₯ plus six tan of π‘₯ plus a constant of integration 𝐢.

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