Integrals of Trig Functions - Online Tutor, Practice Problems & Exam Prep

Integrals Resulting in Basic Trig Functions

So in recent videos, we've been talking about the various rules that you can apply for solving integrals. Now, most of the functions that we've seen up to this point have been these power functions, where we need to apply things like the power rule or the sum and difference rule to integrate them. But it turns out in this course, we're going to come across more of these pesky trig functions that we need to integrate. But thankfully, this process is going to be pretty straightforward if we remember some of the common derivatives we've seen for these trig functions. So without further ado, let's get right into things.

Now just like the process with power functions, finding the integral of a trig function is just going to be the reverse process for finding a derivative. So let's say for example we have the integral of the cosine of x. How could I figure out what this is? Well, something that I know is if I take the derivative of sine, it gives me cosine. So that means the integral of cosine would give me sine.

So the integral of the cosine of x would be the sine of x, plus the constant of integration c. And that right there is the integral of cosine. As you can see, it's pretty straightforward if we just use this reverse logic on the derivatives. Now, we know also the derivative of cosine gives us negative sine. So following this logic, if I took the integral of positive sine, that would give me negative cosine.

So the integral of the positive sine of x would give me the negative cosine of x. Notice how I just took this negative sine and switched it to this other side here, since I was integrating positive sine rather than negative sine. So using this reverse logic, this would be the solution for integrating the sine of x as well. So that's how you can find the integrals of the sine and cosine terms. Now let's actually see if we can use this logic here and apply it to an example where we need to combine multiple of these rules that we've learned about for integrals.

Now here we have the integral of 3 times the sine of x, plus 2 times the cosine of x, all with respect to x. Now the first thing I'm going to do is recognize I can use the sum and difference rule here on these two terms. So what I can do is integrate 3 times the sine of x with respect to x, plus the integral of 2 times the cosine of x with respect to x separately. Now next, I'm going to apply the constant multiple rule, since I can see we have these constants being multiplied by the front of these trig functions. So what I'm gonna do is take that 3 and pull it outside of the integrals.

So I have 3 times the integral of sine of x with respect to x, then I'm going to have plus 2, pulling it outside the integral, times the integral of cosine of x with respect to x. Now from here, I can run the integrals. Now we've seen that the integral of sine gives us negative cosine. So this whole thing is going to become 3 times the negative cosine. As a matter of fact, they'll take that negative sign and write it outside here.

So we have negative 3 times cosine of x plus, and they're going to have 2 right here times this integral. And we see that the integral of cosine comes out to positive sine. So it's plus 2 times the sine of x, and then don't forget the constant of integration c. And that is going to be your solution for this first example, a. Now this was pretty straightforward since we were only dealing with sines and cosines.

What if we had a different example where, say, we were not dealing with sines and cosines, like this example down here? Well, whenever you see these trig functions, we can still use this reverse derivative logic to solve our problem. So in this case, notice we have the integral of 7 secant x tangent x minus cosecant squared x, all with respect to x. Now I'm going to go ahead and use the sum and difference rule again since I can see I can split this into 2 separate integrals. Below the integral of 7 secant x tangent x, and that's going to be minus the integral of cosecant squared x.

Now both of these integrals are going to be with respect to x, since that's what we're integrating here. Now next, I'm going to apply the constant multiple. We want the 7 here and bring it outside of the integral. So we'll have 7 times the integral of secant x tangent x. It's going to be with respect to x minus the integral of cosecant squared x with respect to x.

Now at this point, I need to think about what derivatives would have gotten me to these two functions right here. Starting with secant x tangent x, I know that if I take the derivative of secant x, that's going to give me secant x tangent x. So logically, taking the integral of secant x tangent x would give me just secant x. So we'll get 7 times the secant of x, and that's going to be this first integral right here. And next, I need to subtract off the integral of cosecant squared x.

Now cosecant squared x actually looks familiar to me. So remember that I learned about if I took the derivative of cotangent x, that's going to give me negative cosecant squared x. So likewise, I can use this reverse derivative logic to say that the integral of positive cosecant squared x would give me negative cotangent x. So going down here, we would end up with negative cotangent x. And then don't forget the constant of integration c.

So this would be our integral, which I can clean this up by canceling these negative signs here, giving me 7 secant x plus cotangent x plus c as my solution to this example. So you're going to want to commit these common trig functions to memory when it comes to integration. And you can use the same reverse process for a derivative, so hopefully, it doesn't seem too heavy to memorize. So we know also, for example, that the derivative of tangent gives us secant squared, so we can use this logic to figure out that the integral of secant squared gives us tangent. And another one that we've learned about is we've learned that the derivative of cosecant gives us negative cosecant cotangent.

So that means the integral of cosecant cotangent would give us negative cosecant. So you can always just use this reverse process for derivatives to find the common integrals when dealing with these trig functions. So that is how you can apply these rules and solve these problems, where you have sines and cosines or other more complicated trig functions. Hope you found this video helpful, and let's move on.

Find $g\left(\theta\right)$ by evaluating the following indefinite integral.

$g\left(\theta\right)=\int^{}\left(5\sec^2\theta-2\csc^2\theta\right)d\theta$

• A. $g\left(\theta\right)=5\tan\theta+2\cot\theta+C$
• B. $g\left(\theta\right)=5\tan\theta-2\cot\theta+C$
• C. $g\left(\theta\right)=5\cot\theta+2\tan\theta+C$
• D. $g\left(\theta\right)=5\cot\theta-2\tan\theta+C$

Find $g\left(x\right)$ by evaluating the following indefinite integral.

$g\left(x\right)=\int\left(\sin^2x-100\csc x\cot x+\cos^2x\right)dx$

• A. $g\left(x\right)=100\csc x+x+C$
• B. $g\left(x\right)=-100\csc x+x+C$
• C. $g\left(x\right)=100\csc x+C$
• D. $g\left(x\right)=-100\csc x+C$

Integrals Resulting in Basic Trig Functions Example 1

Here we have an interesting problem where we're asked to verify the following indefinite integral by differentiation. So notice that we're given the indefinite integral of this function right here. This is the integrand. We're integrating this with respect to x, and notice we're also given the solution to this integral. So what are we being asked to do in this problem?

Well, we're asked to verify the following indefinite integral by differentiation. What does this mean? Well, recall that when we do an indefinite integral, this is the same process as taking the antiderivative. We're basically reversing a derivative. So what that means is if this is the solution to this integral, that means we take the derivative of this result that we got, it should give us the function that we have in the integrand right here.

But let's actually verify whether or not this is true. So what I'm going to do is take the derivative of this entire piece of the equation, this entire right side here. So we're going to have the derivative of 2 times the sine of the square root of x. That's going to be plus the constant of integration c. Now, based on the rules for derivatives, sum and difference, I can take this derivative and distribute it to each of these terms.

So we're going to have the derivative of 2 times the sine of the square root of x. We're going to have plus the derivative of c. At this point, there's a couple of things I can do. First off, I can recognize that the derivative of c, this is just a constant, that whole thing's going to go to 0. The derivative of any constant is 0.

Now another thing that I can do is take this 2 that I see here and bring it to the outside of the derivative. So that's going to give me 2 times the derivative of, in this case, we'll have the sine of the square root of x. Now what I need to do at this point is evaluate the derivative of the sine of the square root of x. Doing this, notice we're taking the derivative of sine. The derivative of sine is cosine, so we're going to have 2 times the cosine of the square root of x.

But I can't just finish right here and say that that's my derivative. Because notice that we have a function, the square root of x, within a function, the sine of x. So what I need to do is multiply this whole thing by the derivative of the inside function, and what I see inside here, this function is the square root of x. So we need to multiply this whole thing by the derivative of the square root of x. How exactly could I calculate this derivative right here? Well, we can use power rules to find the derivative here.

Because, see, the derivative of the square root of x, that's the exact same thing as the derivative of x to the one-half power. At this point, I can use the power rule. I can bring that one-half to the front and then reduce the power by 1. Doing this is going to give me one-half \x to the negative one-half power. Anytime you see a negative one-half power, you could write that in the denominator of a fraction and make that exponent positive.

So, we'll have one-half, and that's going to be multiplied by 1 over x to the one-half power. And I could rewrite this whole thing to be 1 over 2 times the square root of x. This is what we get for the derivative of x to the one-half power, or the square root of x. So going back down here, what we're going to have is 2 times the cosine of the square root of x. That's going to be multiplied by the derivative of the square root of x, which we just set up here as 1 over 2 times the square root of x.

Now, at this point, I can cancel this 2 with that 2 right there. So all I'm going to end up with, which I'll write it up here, is the cosine of the square root of x multiplied by 1 over the square root of x, which you can just take this one over square root of x right here and write it as a denominator to what we have there. So this is what we end up giving as our result. And this right here would be the derivative of the right side of this equation. So the derivative of this entire right side here is what we already calculated.

We have 2 times the sine of the square root of x plus c. What we got down here is going to be the same results, which is the derivative of this whole thing right here. We would get this as our result. And that's how you can solve this problem, and notice something about the result that we got. Notice that the result that we got is the exact same thing as what's in the integrand. So this cosine square root x over square root x is the same result that we got when we took this derivative.

So, as you can see, since these results match when taking the derivative of this right side, we have verified that this integral is correct. So that is how you can verify an integral by using differentiation. All you want to do is differentiate the result to see if it comes out to be the same thing as what you have in the integrand, since taking the integral is really just the reverse process of differentiation. I hope you found this video helpful, and let's move on.