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

Determine ∫ (βˆ’6 sin 7π‘₯ + 2 secΒ² 6π‘₯ βˆ’ 1) dπ‘₯.

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Video Transcript

Determine the integral of negative six sin of seven π‘₯ plus two times the sec squared of six π‘₯ minus one with respect to π‘₯.

In this question, we’re asked to evaluate the integral of three separate terms. And two of these terms are trigonometric functions, and the third term is a constant. We know how to evaluate the integral of all three terms individually. Therefore, by using the properties of integrals, we can just integrate each term individually. This gives us the integral of negative six sin of seven π‘₯ with respect to π‘₯ plus the integral of two sec squared of six π‘₯ with respect to π‘₯ plus the integral of negative one with respect to π‘₯.

Before we evaluate each integral individually, let’s start by taking a constant factor outside of each of our integrals. We take the factor of negative six out of our first integral, the factor of two out of our second integral, and the factor of negative one outside of our third integral to get the following expression. We’re now ready to evaluate each of our integrals separately. Let’s start with our first integral.

We need to integrate the sine function. And to do this, we recall the following result. For any real constant π‘Ž not equal to zero, the integral of the sin of π‘Žπ‘₯ with respect to π‘₯ is equal to negative one over π‘Ž multiplied by the cos of π‘Žπ‘₯ plus the constant of integration 𝐢. In our function, the value of π‘Ž is equal to seven. Substituting seven into our result, we get negative one over seven multiplied by the cos of seven π‘₯. And although we can add the constant of integration for each of our integrals, we would then combine this at the end. So it’s easier just to add one constant of integration at the end of our expression anyway. Finally, we need to multiply our expression by the factor outside of our integral. This gives us negative one times negative six over seven multiplied by the cos of seven π‘₯.

Let’s now evaluate our second integral, the integral of the sec squared of six π‘₯ with respect to π‘₯. And to do this, we recall the following result. For any real constant π‘Ž not equal to zero, the integral of the sec squared of π‘Žπ‘₯ with respect to π‘₯ is equal to one over π‘Ž multiplied by the tan of π‘Žπ‘₯ plus the constant of integration 𝐢. This time, our value of π‘Ž is equal to six. And remember, we have a coefficient of two outside of our integral. Substituting π‘Ž is equal to six into this result and multiplying by two, we get two times one-sixth tan of six π‘₯.

Finally, we need to evaluate our final integral, the integral of one with respect to π‘₯. There’s a few different ways of doing this. For example, we could use the power rule for integration. However, it’s easier just to remember the derivative of any linear function is the coefficient of π‘₯. So the derivative of π‘₯ with respect to π‘₯ is one. This means π‘₯ is an antiderivative of one. So when we integrate a constant, we just multiply it by π‘₯. Remember, we’re subtracting this value, so we need to subtract our value of π‘₯. And finally, remember we do need to add a constant of integration at the end of our expression.

Then, all we need to do is add all of these terms together to get an expression for our integral. We can then simplify our coefficients. We get six over seven times cos of seven π‘₯ plus one-third tan of six π‘₯ minus π‘₯ plus our constant of integration 𝐢. Finally, we’ll reorder our terms by bringing the term negative π‘₯ to the front of our expression. This then gives us our final answer. We were able to show the integral of negative six sin of seven π‘₯ plus two sec squared of six π‘₯ minus one with respect to π‘₯ is equal to negative π‘₯ plus six over seven times the cos of seven π‘₯ plus one-third tan of six π‘₯ plus a constant of integration 𝐢.

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