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.

16:32

Video Transcript

In this video, we’ll learn how to find indefinite integrals of trigonometric functions. We’ll begin by recalling what we mean by an antiderivative before looking at how this helps us to integrate a number of trigonometric functions.

Let’s start by recalling what the antiderivative of a function is. Uppercase 𝐹 is the antiderivative of lowercase 𝑓 if uppercase 𝐹 prime of π‘₯ is equal to lowercase 𝑓 of π‘₯. And that’s where uppercase 𝐹 prime of π‘₯ is the derivative with respect to π‘₯ of uppercase 𝐹 of π‘₯. In fact, we can say this is true for any function 𝐺 of π‘₯, where 𝐺 of π‘₯ is equal to uppercase 𝐹 of π‘₯ plus 𝑐, for any constant 𝑐. Now this is very useful as we’ll be using it to define an indefinite integral.

We can say that the indefinite integral of lowercase 𝑓 of π‘₯ with respect to π‘₯ is equal to uppercase 𝐹 of π‘₯ plus 𝑐, where uppercase 𝐹 is the antiderivative of lowercase 𝑓. And it’s very important to remember our constant of integration 𝑐 when performing an indefinite integral. And note that it’s called an indefinite integral as there are no limits of integration specified. That is, we’re not integrating over a specific interval of π‘₯-values as we would be in a definite integral. And this is a little beyond the scope of this video since it touches on the fundamental theorem of calculus.

However, we can see why the constant 𝑐 is necessary if we perform the reverse operation on our indefinite integral equation. So that’s differentiating with respect to π‘₯. And since we’re simply performing the inverse operation on the left, the derivative with respect to π‘₯ of the integral of lowercase 𝑓 of π‘₯ is just lowercase 𝑓 of π‘₯. And on the right-hand side, when we differentiate uppercase 𝐹 of π‘₯ with respect to π‘₯, we simply get 𝐹 prime of π‘₯. And our constant 𝑐 will simply vanish because differentiating a constant gives zero. And now when we go back the other way to our integral, the constant will appear again. But since we don’t know the value of this constant, we simply label it 𝑐. It’s just an unknown constant.

We can use this idea of the integral as antiderivative together with our knowledge of the derivatives of various trigonometric functions to find some specific integrals. So let’s begin by looking at the function 𝑓 of π‘₯ is equal to the sin of π‘Žπ‘₯, for real constants π‘Ž and where for this to be valid π‘₯ must be in radian measure. And we want to find the indefinite integral of 𝑓 of π‘₯ evaluated with respect to π‘₯. We recall that the derivative of the cos of π‘Žπ‘₯ with respect to π‘₯ is equal to negative π‘Ž times the 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 the cos of π‘Žπ‘₯. Don’t forget though that since we’re working with an indefinite integral, we need to add a constant of integration, which we’ll call lowercase 𝑐.

Okay, so this is good start. But we’re actually looking to find the indefinite integral of the sin of π‘Žπ‘₯, not negative π‘Ž times the sin of π‘Žπ‘₯. We’re allowed though to take the constant, negative π‘Ž, outside of the integral. And so we see that negative π‘Ž times the indefinite integral of sin π‘Žπ‘₯ is equal to the cos of π‘Žπ‘₯ plus the constant 𝑐. And since negative π‘Ž is a constant, we can divide both sides by negative π‘Ž. And we obtain the indefinite integral of sin π‘Žπ‘₯ to be negative one over π‘Ž times the cos of π‘Žπ‘₯ plus uppercase 𝐢. And notice that we’ve changed from lowercase 𝑐 to uppercase 𝐢 because we’ve divided our original constant by another constant, negative π‘Ž, and we need to represent this change in value.

Making a note of our first result for the indefinite integral of sin π‘Žπ‘₯, we can repeat this process for the function 𝑓 of π‘₯ is equal to cos π‘Žπ‘₯, where again π‘Ž is a real constant and π‘₯ is measured in radians. We’re going to use the fact that the derivative with respect to π‘₯ of sin π‘Žπ‘₯ is equal to π‘Ž times the cos of π‘Žπ‘₯. And we can therefore say that the indefinite integral of π‘Ž cos π‘Žπ‘₯ evaluated with respect to π‘₯ is equal to sin π‘Žπ‘₯ plus the constant of integration lowercase 𝑐. And again, we’re allowed to take the constant π‘Ž outside the integral, and the right-hand side remains unchanged. And finally, we divide through by π‘Ž and obtain the indefinite integral of cos π‘Žπ‘₯ evaluated with respect to π‘₯ to be one over π‘Ž times the sin of π‘Žπ‘₯ plus the new constant of integration uppercase 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 equal to cos π‘₯, where here our constant π‘Ž is equal to one. Differentiating again gives us negative sin π‘₯. And differentiating again, we get negative cos π‘₯. And differentiating one more time, we go back to our original function, sin π‘₯. We reverse this cycle as shown for integration. Let’s now look at some examples illustrating the integration of sine and cosine functions.

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

Before trying to evaluate this, it can be useful to recall some of the properties of integrals. Firstly, the integral of the sum of two or more functions is 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. These properties mean 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‘₯.

And now we recall the general results for the integrals of the sine and cosine functions. The indefinite integral of sin of π‘Žπ‘₯ is equal to negative one over π‘Ž times cos π‘Žπ‘₯ plus the constant of integration 𝑐. And the integral of cos of π‘Žπ‘₯ evaluated with respect to π‘₯ is equal to one over π‘Ž times sin π‘Žπ‘₯ plus the constant 𝑐. So in our case, the constant π‘Ž is equal to one, and our integral is negative negative cos π‘₯ plus the constant 𝐴 minus nine times sin π‘₯ plus the constant 𝐡. And we’ve chosen 𝐴 and 𝐡 to show that these are different constants of integration.

Now distributing the parentheses and combining the two constants 𝐴 and 𝐡 into a single constant 𝐢, we have the indefinite integral of negative sin π‘₯ minus nine cos π‘₯ evaluated with respect to π‘₯ is equal to cos π‘₯ minus nine times sin π‘₯ plus the constant 𝐢.

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

In this question, we’re looking to integrate the sum of two functions of π‘₯. We begin by recalling that the integral of the sum of two or more functions is equal to the sum of the integrals of those respective functions. And so we can write our integral 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 integrals and focus on integrating each expression in π‘₯. And this means we can rewrite our integrals as negative eight times the integral with respect to π‘₯ of sin eight π‘₯ minus seven times the integral of cos five π‘₯ dπ‘₯.

Next, we recall the general results for the integrals of sin π‘Žπ‘₯ and cos π‘Žπ‘₯. The indefinite integral with respect to π‘₯ of sin of π‘Žπ‘₯ is equal to negative one over π‘Ž times cos π‘Žπ‘₯ plus the constant of integration 𝑐. And the indefinite integral of cos π‘Žπ‘₯ with respect to π‘₯ is one over π‘Ž sin π‘Žπ‘₯ plus the constant 𝑐. In our case, the constant π‘Ž is eight in the first integral and five in the second integral. And applying these results to our integrals, we see that the integral of sin of eight π‘₯ is negative one over eight cos eight π‘₯ plus the constant 𝐴. And the integral with respect to π‘₯ of cos of five π‘₯ is one over five times the sin of five π‘₯ plus 𝐡. Note that we’ve chosen 𝐴 and 𝐡 as the constant of integration, not just a single value of 𝑐, to show that these are actually different constants.

Our final step is to distribute the parentheses. Negative eight times negative one over eight. cos of eight π‘₯ is just the cos of eight π‘₯. And negative seven times one over five sin five π‘₯ is negative seven over five sin five π‘₯. And finally, we multiply negative eight by 𝐴 and negative seven by 𝐡. And since we don’t know the values of 𝐴 and 𝐡, we choose to represent this as a single constant 𝑐. We found then that the integral we required is the cos of eight π‘₯ minus seven over five times the sin of five π‘₯ plus the constant 𝑐.

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 this video for a moment to consider what this tells us about the integral of sec squared π‘Žπ‘₯. Well, let’s have a look. We recall that an integral can be considered as the antiderivative. That is, integration is the reverse process for differentiation. And we can see then that the indefinite integral of π‘Ž sec squared π‘Žπ‘₯ with respect to π‘₯ must be tan of π‘Žπ‘₯ plus the constant of integration 𝑐. We take out the constant factor of π‘Ž. And then we divide through by π‘Ž. Then we can see that the integral of sec squared π‘Žπ‘₯ with respect to π‘₯ is one over π‘Ž tan of π‘Žπ‘₯ plus uppercase 𝐢.

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‘₯ with respect to π‘₯ is negative one over π‘Ž times csc π‘Žπ‘₯ plus 𝑐. The integral of sec π‘Žπ‘₯ tan π‘Žπ‘₯ with respect to π‘₯ is one over π‘Ž sec π‘Žπ‘₯ plus 𝑐. And the integral of csc squared π‘Žπ‘₯ with respect to π‘₯ negative one over π‘Ž times cot of π‘Žπ‘₯ plus 𝑐. We’re now going to look at some examples of these results and how often we use trigonometric identities to help us evaluate these integral.

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

To answer this question, it’s almost simply enough to quote the general result for the integral of sec squared π‘Žπ‘₯, where in our case the constant π‘Ž is equal to six. The result tells us that the integral of sec squared π‘Žπ‘₯ with respect to π‘₯ is one over π‘Ž times the tan of π‘Žπ‘₯ plus constant 𝑐. It’s sensible, however, before using this result to take the factor of negative one outside the integral as shown. So when we do perform our integration, we obtain the solution to be negative one times one-sixth of tan of six π‘₯ plus 𝑐, recalling that in our case the constant π‘Ž is equal to six.

And now all that’s left is to distribute the parentheses. Negative one times one-sixth of tan six π‘₯ is negative one-sixth tan of six π‘₯. And negative one times the constant 𝑐 gives us this new constant uppercase 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 of three π‘₯ plus one over nine cos squared of three π‘₯ with respect to π‘₯.

Now at first glance, this integral looks quite complicated. We should spot, however, that we can actually simplify the quotient. How we do this is to essentially reverse the process we follow 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‘₯. By dividing through in the first fraction by cos squared three π‘₯, our first term simplifies to two over nine cos three π‘₯. And then to help us to spot what to do next, let’s rewrite the second fraction as a ninth times one over cos squared of three π‘₯.

Next, we recall that the integral of the sum of two or more functions is the sum of their integrals. And we also know that we can take any constant factors outside of the integral and focus on integrating the expression in π‘₯. And applying these properties gives us two over nine times the integral of the cos of three π‘₯ with respect to π‘₯ plus one over nine times the integral of one over the cos squared of three π‘₯ again with respect to π‘₯. We can quote the result for the integral of cos of π‘Žπ‘₯. It’s one over π‘Ž times the sin of π‘Žπ‘₯ plus the constant of integration 𝑐. And this means that the integral of cos of three π‘₯ is a third sin of three π‘₯ plus some constant of integration which we’ll call 𝐴.

But what do we do about the second integral? Well, we know the trigonometric identity one over cos of π‘₯ is equal to sec π‘₯. And so we see that we can rewrite one over cos squared of three π‘₯ as sec squared of three π‘₯. So now we can use the general result for the integral of sec squared of π‘Žπ‘₯ with respect to π‘₯. And that’s one over π‘Ž times the tan of π‘Žπ‘₯ plus some constant 𝑐. And this means we can write the integral of sec squared three π‘₯ with respect to π‘₯ as one over three tan three π‘₯ plus a constant of integration which we’ll call 𝐡.

Finally, we distribute our parentheses and we see that two over nine times one-third of sin three π‘₯ is equal to two over 27 times sin three π‘₯. And one over nine times one over three tan three π‘₯ is equal to one over 27 tan three π‘₯. We multiply each of our constants by two over nine and one over nine, respectively. And we end up with a new constant 𝐢. We find then that our integral is two over 27 times the sin of three π‘₯ plus one over 27 times tan of three π‘₯ plus the constant of integration 𝐢.

Let’s consider one more example that requires the integrals we’ve looked at together with some trigonometric identities.

Determine the indefinite integral of negative three tan squared of eight π‘₯ times csc squared eight π‘₯ evaluated with respect to π‘₯.

This does at first look quite tricky. However, if we recall some of our trigonometric identities, we can change this into something more manageable. We know that tan π‘₯ is equal to sin π‘₯ over cos π‘₯ and that csc π‘₯ is equal to one over six π‘₯. We can therefore rewrite our integrand, that’s the function we want to integrate, as negative three times sin squared of eight π‘₯ over cos squared of eight π‘₯ times one over sin squared of eight π‘₯. And then we see that we can cancel the sin squared of eight π‘₯. We can take the factor of negative three outside our integral to make the next step easier. And we have negative three times the integral of one over cos squared eight π‘₯ evaluated with respect to π‘₯.

Now we know that one over cos π‘₯ is equal to sec π‘₯. So our integral becomes negative three times the integral of sec squared of eight π‘₯ with respect to π‘₯. But of course the integral of sec squared π‘Žπ‘₯ with respect to π‘₯ is one over π‘Ž tan of π‘Žπ‘₯ plus constant 𝑐. And so in our case where the constant π‘Ž is equal to eight, the integral of sec squared eight π‘₯ with respect to π‘₯ is one over eight times the tan of eight π‘₯ plus the constant 𝑐. We distribute our parentheses and we see that our answer is negative three-eighths times tan of eight π‘₯ plus a new constant, since we multiplied our original one by negative three. So let’s call that uppercase 𝐢.

Let’s now complete this video by recalling some of the key points we’ve covered. 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 identities such as tan equals sin π‘₯ over cos π‘₯ or one over sin π‘₯ is equal to cos of π‘₯ can make integrals easier to evaluate.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.