Lesson Video: Indefinite Integrals: Trigonometric Functions Mathematics • Higher Education

In this video, we will learn how to find indefinite integrals that result in trigonometric functions.

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Video Transcript

In this video, we’ll learn how to find indefinite integrals of trigonometric functions. We’ll begin by recalling the first part of the fundamental theorem of calculus before looking at how this helps us to integrate a number of trigonometric functions and the applications of these integrals. We begin by stating the first part of the fundamental theorem of calculus. In this theorem, we let 𝑓 be a continuous real valued function defined on a closed interval π‘Ž to 𝑏. Then capital 𝐹 is the function defined for all π‘₯ in this closed interval by capital 𝐹 of π‘₯ equals the integral of 𝑓 of 𝑑 with respect to 𝑑 evaluated between π‘Ž and π‘₯. Then capital 𝐹 must be uniformly continuous on this closed interval and differentiable on the open interval π‘Ž to 𝑏 such that capital 𝐹 prime of π‘₯ is equal to 𝑓 of π‘₯ for all π‘₯ in the closed interval π‘Ž to 𝑏.

This last part says 𝐹 is the antiderivative of the function 𝑓. The function whose derivative is equal to the original function. And essentially, what this tells us is that integration is the reverse process, the differentiation. So let’s begin by looking at the function 𝑓 of π‘₯ equals sin of π‘Žπ‘₯ for real constants π‘Ž. And of course, π‘₯ must be a radiant measure. We recall that the derivative of cos of π‘Žπ‘₯ is negative π‘Ž sin of π‘Žπ‘₯. We can, therefore, say that the indefinite integral of negative π‘Ž sin of π‘Žπ‘₯ evaluated with respect to π‘₯ must be cos of π‘₯. And don’t forget that since we’re working with an indefinite integral, we need to add this constant of integration; let’s call that 𝑐.

Okay, this is a great start. But we’re actually looking to find the indefinite integral of sin of π‘Žπ‘₯ not negative π‘Ž sin of π‘Žπ‘₯. We’re allowed, though, to take the constant negative π‘Ž outside of the integral. And we see that negative π‘Ž times the indefinite integral of sin of π‘Žπ‘₯ is equal to cos of π‘Žπ‘₯ plus 𝑐. And since negative π‘Ž is just a constant, we divide both sides by negative π‘Ž. And we obtain the indefinite integral of sin of π‘Žπ‘₯ to be negative one over π‘Ž times cos of π‘Žπ‘₯ plus capital 𝐢. And you might notice that I’ve changed from a lowercase 𝑐 to an uppercase 𝐢. And this is simply because we’ve divided our original constant by another constant. So I want to represent that this number has actually changed in value.

We can repeat this process for the function 𝑓 of π‘₯ equals cos π‘Ž of π‘₯. Once again, π‘Ž is a real constant and π‘₯ is measured in radians. We’re going to use the fact that the derivative of sin of π‘Žπ‘₯ is equal to π‘Ž cos of π‘Žπ‘₯. And we can, therefore, say that the indefinite integral of π‘Ž cos of π‘Žπ‘₯ evaluated with respect to π‘₯ must be sin π‘Žπ‘₯ plus the constant of integration 𝑐. We take the constant π‘Ž outside of the integral, and the right-hand side of our equation remains unchanged. Finally, we divide through by π‘Ž. And we obtain the integral of cos of π‘Žπ‘₯ evaluated with respect to π‘₯, to be one over π‘Ž times sin of π‘Žπ‘₯ plus capital 𝐢.

Now you might recall that the process for differentiating sine and cosine functions forms a cycle. That is, the derivative of sin π‘₯ is cos π‘₯. And if we differentiate again, we get negative sin π‘₯. Differentiating again, we get negative cos of π‘₯. And differentiating one more time, we go back to the start and get sin of π‘₯. We reverse this cycle for integration as shown. Let’s now look at some examples demonstrating the integration of the sin and cos functions.

Determine the indefinite integral of negative sin of π‘₯ minus nine cos of π‘₯ evaluated with respect to π‘₯.

It can be useful to recall the properties of integrals before evaluating this. Firstly, we know 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 also know that we can take any constant factors outside of the integral and focus on integrating the expression in π‘₯ itself. This means we can rewrite our integral as negative the integral of sin π‘₯ evaluated with respect to π‘₯ minus nine times the integral of cos of π‘₯ evaluated with respect to π‘₯. This means we can rewrite our integral as shown. And this means we can recall the general results for the integral of the sine and cosine functions.

The indefinite integral of sin of π‘Žπ‘₯ is negative one over π‘Ž cos π‘Žπ‘₯ plus that constant 𝑐. And the indefinite integral of cos of π‘Žπ‘₯ is one over π‘Ž times sin of π‘Žπ‘₯ plus 𝑐. So our integral is negative negative cos of π‘₯ plus 𝐴 minus nine times sin of π‘₯ plus 𝐡. Then I’ve chosen 𝐴 and 𝐡 here to show that these are different constants of integration. Distributing the parentheses and combining our constants in to one constant, we find the integral of negative sin π‘₯ minus nine cos of π‘₯ evaluated with respect to π‘₯ to be cos of π‘₯ minus nine sin of π‘₯ plus 𝑐.

Determine the indefinite integral of negative eight sin of eight π‘₯ minus seven cos of five π‘₯ evaluated with respect to π‘₯.

In this question, we’re looking to integrate the sum of two functions in π‘₯. We begin by recalling the fact that the integral of the sum of two or more functions is actually equal to the sum of the integrals of those respective functions. So we can write our problem as the integral of negative eight sin of eight π‘₯ with respect to π‘₯ plus the integral of negative seven cos of five π‘₯ dπ‘₯. We also know that we can take any constant factors outside of the integral and focus on integrating each expression in π‘₯. So we can rewrite our problem further as negative eight times the integral of sin of eight π‘₯ with respect to π‘₯ minus seven times the integral of cos of five π‘₯ with respect to π‘₯.

Next, we recall the general results for the integral of sine and cosine. The integral of sin of π‘Žπ‘₯ is negative one over π‘Ž cos of π‘Žπ‘₯ plus 𝑐. And the indefinite integral of cos of π‘Žπ‘₯ is one over π‘Ž sin of π‘Žπ‘₯ plus 𝑐. We integrate each function, respectively, and we see that the integral of sin of eight π‘₯ is negative one-eighths cos of eight π‘₯ plus 𝐴. And the indefinite integral of cos of five π‘₯ is a fifth sin of five π‘₯ plus 𝐡. And I’ve chosen 𝐴 and 𝐡, as opposed to just one value of 𝑐, to show that these are actually different constants.

Our final step is to distribute the parentheses. Negative eight times negative one-eighths cos of eight π‘₯ is just cos of eight π‘₯. Negative seven times one-fifth of sin of five π‘₯ is a negative seven-fifths sin of five π‘₯. Finally, we multiply negative eight by 𝐴 and negative seven by 𝐡. And we end up with this new constant 𝐢. And we found that the integral we required is cos of eight π‘₯ minus seven-fifths of sin of five π‘₯ plus 𝐢.

We’re now going to consider some alternative derivatives. We recall that the derivative of tan π‘Žπ‘₯ is π‘Ž sec squared π‘Žπ‘₯. Now, you might like to pause the video for a moment and consider what this tells us about the integral of sec squared π‘Žπ‘₯. Well, let’s have a look. We recall that the first part of the fundamental theorem of calculus essentially tells us that integration is the reverse process for differentiation. And we can see then that the integral of π‘Ž sec squared π‘Žπ‘₯ evaluated with respect to π‘₯ must be tan of π‘Žπ‘₯ and since this is an indefinite integral plus 𝑐. We take out this constant factor of π‘Ž and then we divide. And we see that the integral of sec squared of π‘Žπ‘₯ evaluated with respect to π‘₯ is one over π‘Ž tan of π‘Žπ‘₯ plus capital 𝐢.

It is outside of the scope of this video to spend too long looking at the other integrals. But using much the same method, we obtain the following integrals relating to the derivatives of the reciprocal trigonometric functions. The integral of csc π‘Žπ‘₯ cot π‘Žπ‘₯ is negative one over π‘Ž csc π‘Žπ‘₯. The integral of sec π‘Žπ‘₯ tan π‘Žπ‘₯ is one over π‘Ž sec π‘Žπ‘₯. And the derivative of csc squared π‘Žπ‘₯ is negative one over π‘Ž times cot of π‘Žπ‘₯. We’re now going to have a look at some of the examples of these results and how often times the trigonometric identity will be required to help us evaluate these integrals.

Determine the indefinite integral of negative sec squared six π‘₯ evaluated with respect to π‘₯.

To answer this question, it’s almost simply enough to quote the general result for the integral of sec squared of π‘Žπ‘₯. It’s one over π‘Ž times the tan of π‘Žπ‘₯. It is sensible, though, to take a factor of negative one outside of the integral as shown. And when we do, we obtain the solution to be negative one times a sixth of tan six π‘₯ plus 𝑐. And all that’s left is to distribute the parentheses. Negative one times a sixth of tan six π‘₯ is negative one-sixth tan of six π‘₯. And negative one times 𝑐 gives us this new constant, capital 𝐢. And so we find our indefinite integral to be negative one sixth of tan of six π‘₯ plus 𝐢.

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.

We’ll consider one more example that requires the integrals that we’ve looked at plus some trigonometric identities.

Determine the indefinite integral of negative three tan squared eight π‘₯ times csc squared eight π‘₯ evaluated with respect to π‘₯. These does at first look quite tricky. However, If we recall some of our trigonometry identities, it does get a little nicer. We know that tan π‘₯ is equal to sin π‘₯ over cos of π‘₯. And we also know that csc π‘₯ is equal to one over sin π‘₯. We can, therefore, rewrite our entire integrant as negative three times sin squared eight π‘₯ over cos squared eight π‘₯ times one over sin squared eight π‘₯. And then we noticed that the sin squared eight π‘₯ cancels. We can take the factor of negative three outside of the integral sin to make the next step easier. And we have a negative three times the integral of one over cos squared eight π‘₯ dπ‘₯.

But we know that one over cos of π‘₯ is equal to sec of π‘₯. So our integral becomes negative three times the integral of sec squared eight π‘₯ evaluated with respect to π‘₯. But of course, the integral of sec squared π‘Žπ‘₯, evaluated with respect to π‘₯, is one over π‘Ž tan of π‘Žπ‘₯ plus some constant of integration 𝑐. And so we see that the integral of sec squared eight π‘₯ is an eighth tan of eight π‘₯ plus 𝑐. We distribute our parentheses. And we see that we’re left with negative three-eights of tan of eight π‘₯ plus a new constant since we multiplied our original one by negative three. Let’s call that capital 𝐢.

In this video, we’ve seen that we can use the fact that integration is essentially the reverse process of differentiation to evaluate the indefinite integrals of sin of π‘Žπ‘₯, cos of π‘Žπ‘₯, and sec squared of π‘Žπ‘₯. We’ve also seen that recalling certain trigonometric identity, such as tan π‘₯ equals sin π‘₯ over cos π‘₯ or one over sin π‘₯ is equal to csc of π‘₯, can make integrals easier to evaluate.

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