Evaluate the indefinite integral.∫x2cos4xdx\int_{}^{}x^2\cos4xdx

x 2 4 sin ⁡ ( 4 x ) + 1 4 cos ⁡ ( 4 x ) − 1 16 sin ⁡ ( 4 x ) + C \frac{x^2}{4}\sin\left(4x\right)+\frac14\cos\left(4x\right)-\frac{1}{16}\sin\left(4x\right)+C

Evaluate the indefinite integral. ∫ x 2 cos 4 x d x \int_{}^{}x^2\cos4xdx

this problem, we are asked to evaluate the indefinite integral. And notice here we have the product of two functions, x squared and cosine of 4x. How do we solve this problem? Now, when we have the product of two functions, a good place to start is oftentimes to see if we can do a variable substitution anywhere. Now, I could try to do a variable substitution on the 4x, but if I take the derivative of 4x, it would just give me four if I take that derivative with respect to x, and that's not going to help me get rid of this x squared in front. Now if I did a variable substitution on the x squared, well that's going to give me the derivative of 2x, but that's not going to allow me to clear the 4x inside the cosine. So this is a situation where we're gonna need to use integration by parts. Now for integration by parts in this case, what I'm going to do is create a table. So creating a table here, we're going to have one column for our derivatives and another for our integrals. Now you may recall that the derivatives, that's just gonna be whatever our u is, and then the integrals is going to be whatever the dv is. So first off, let's start with the derivatives. Well, what's gonna go in my derivative column is something that can that we can get more simple if we take the derivative. And looking at this, if I were to set my cosine equal to u or put it in my derivative column, that cosine, when we take the derivative, would give us negative sign. And it's also going to have this this four on the outside here because of the chain rule. And then we just kinda get this back and forth argument between the sine and cosine as we kept taking the derivative. That's not really going to become more simple at any point. But if we took the derivative of x squared, well, that would give me two x. So we go from an x squared term or a term with x to an exponent raised to an exponent of two to an exponent of one. That is gonna be something more helpful. So I'm going to say that this derivative column is going to have x squared. And I can use the power rule and take the derivative of x squared, and that would give me two x. I could go ahead and take the derivative of two x, and that would just give me two. Can take the derivative of two, and that's gonna go ahead and give me zero. And we'll go ahead and stop right about there because I don't think we can get a lot more simple than zero. So now let's move on to our integral column. Our integral column is going to be the dv, and keep in mind this x squared is what we said equal to u. Since we set x squared equal to u, it makes sense that the dv is going to be this cosine, because the cosine of 4x, it needs to be something that can be easily integrated, and I know how to integrate sine and cosine functions. So I'm going to say that my dv here is going to be the cosine of 4x. Now Now how can we go about integrating this? Well if you were to integrate the cosine of four x you would need to do some variable substitution to solve this problem, but the way you could do it is by taking this four x and setting that equal to u. So u would be equal to four x and the derivative of u would be four with the respect to x. So what that means is we could write this whole thing as the integral of the cosine of u, and then it's gonna be dx, but notice I could take that dx and I could multiply it by four. So so say four times dx and I'll go ahead and divide by four too just to go ahead and get the same argument that we had originally, and that will allow me to substitute du in right there. So we end up with one fourth and it's going to be times the cosine of u times the integral of the cosine of u du, replacing this with a du, and then integrating the cosine will give you sine. So you get one fourth, that's gonna be the sine of u plus c. Then I can go ahead and take that u, and I could replace it with what we had in the first place for u which was four x. So we get one fourth times the sine of four x plus c. So integrating this is going to give us one fourth times the sine of four x. So that's what we get. Now going through this process, you could go ahead and just integrate this and go through the same situation with the variable substitution, but eventually you're gonna start to notice a pattern here. Notice we're just taking whatever is in front of the x here and we're dividing it on the outside. And in addition, we're applying the same rules that we know for integrating trig functions, so like the cosine got integrated to become a sine. Now I know the integral of sine is negative cosine, so I'm going to get negative and then bring this four into the denominator. Four times four will give us one eighth, then we're going to go back to cosine, that's going to be four x integrating again. If I were to integrate this again, well, what I'm going to do is once again I I know that the integral of cosine will just be sine, and so what that's going to give me is negative, and we'll have one over, in this case, we'll have four times eight, which is 32. So I have one over 32. That's going to be sine of four x, and that's where we're gonna go ahead and stop. So a lot of times, that's what's really nice with doing this pattern with the tables, is eventually you can start to recognize a pattern with your integrals, and you can go ahead and just keep applying that pattern until you get to however many integrals or derivatives that you want. Now one more step you wanna make sure you do is alternating the signs on the derivative column. You should have positive, then negative, then positive, then negative, and that's where we'll go ahead and stop. So now we have our table filled out here, and so we can go ahead and apply this to solve the problem. Now first off, what you wanna do is start with whatever you have here in your derivative column, and then you wanna move diagonally down into the right. So we're going to get x squared times, and then it's going to be one fourth sine of 4x. Now next what I'm going to do is move to this piece, I'm going to go ahead and go down into the right once again diagonally. So it's going to be negative two x times negative one eighth cosine of four x, but notice here we do with these two negative signs. So those will cancel and just give us a positive. So I have positive, and it's going to be two over eight, and that's going to be cosine four x. So we could reduce this fraction, but I'm gonna do that in a moment here because I wanna simplify everything at once here. So this is what we're getting so far, and then we're going to have plus, I'm just gonna write it down here, plus, and then we're gonna move to this piece. We're gonna multiply that two down there. That's going to give us two over 32. That's gonna be sine of four x, and I notice here that we actually have a negative sign there and a positive there which could give us a negative, so this right here should actually be a minus right here, so it's gonna be minus this. And then finally lastly I'll move to this one plus, and then we're going to multiply straight across here. But notice since we're going straight across rather than diagonally down, we have to go ahead and do an integral. So I'm going to have negative zero times this, and zero times anything is just zero, so we don't even need to worry about that integral right there. We're just gonna get the plus c at the end. So So this is what we end up getting. Now from here what I can do is simplify. So I'm gonna get an x squared over four up front here that's gonna be sine 4x. Now next what I'm gonna have is plus, and that's going to be two over eight, which reduces to one fourth, and then we have cosine 4x, and we're going to have minus, and then we have two over 32, which that's going to reduce to one sixteenth, so we'll have one sixteenth, and then we'll have sine four x. And then we'll have plus the constant integration c, and that right there is the solution to the problem. So this is how you can solve repeated integration by parts problems where you use a table instead of just trying to brute force this with the formula. I mean, you could get the same answer with using the formula, but oftentimes writing a table can help you to organize your terms better and get to a result a little more quick. So hope you found this video helpful. Let me know if you have any questions, and let's move on.

Evaluate the indefinite integral. ∫ x 2 cos ⁡ 4 x d x \int_{}^{}x^2\cos4xdx

x 2 4 sin ⁡ ( 4 x ) + 1 4 cos ⁡ ( 4 x ) − 1 16 sin ⁡ ( 4 x ) + C \frac{x^2}{4}\sin\left(4x\right)+\frac14\cos\left(4x\right)-\frac{1}{16}\sin\left(4x\right)+C

x 2 4 sin ⁡ ( 4 x ) + 1 4 cos ⁡ ( 4 x ) − 1 16 cos ⁡ ( 4 x ) + C \frac{x^2}{4}\sin\left(4x\right)+\frac14\cos\left(4x\right)-\frac{1}{16}\cos\left(4x\right)+C

x 3 3 cos ⁡ ( 4 x ) + 4 x 2 sin ⁡ ( 4 x ) + C \frac{x^3}{3}\cos\left(4x\right)+4x^2\sin\left(4x\right)+C

4 x 3 3 cos ⁡ ( 4 x ) + 4 x 2 sin ⁡ ( 4 x ) + C \frac{4x^3}{3}\cos\left(4x\right)+4x^2\sin\left(4x\right)+C

12. Techniques of Integration

Integration by Parts

Calculus

12. Techniques of Integration

Integration by Parts

Struggling with Calculus?

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Multiple Choice

Evaluate the indefinite integral.
$\int_{}^{} x^{2} \cos 4 x d x$

$\frac{x^{2}}{4} \sin (4 x) + \frac{1}{4} \cos (4 x) - \frac{1}{16} \cos (4 x) + C$

$\frac{x^{3}}{3} \cos (4 x) + 4 x^{2} \sin (4 x) + C$

$\frac{x^{2}}{4} \sin (4 x) + \frac{1}{4} \cos (4 x) - \frac{1}{16} \sin (4 x) + C$

$\frac{4 x^{3}}{3} \cos (4 x) + 4 x^{2} \sin (4 x) + C$

Verified step by step guidance

Step 1: Recognize that the integral ∫x^2cos(4x)dx requires integration by parts. Recall the formula for integration by parts: ∫u dv = uv - ∫v du. Here, we will choose u = x^2 and dv = cos(4x)dx.

Step 2: Differentiate u and integrate dv. Compute du = d(x^2) = 2x dx, and integrate dv to get v = ∫cos(4x)dx = (1/4)sin(4x).

Step 3: Substitute into the integration by parts formula. Using the formula ∫u dv = uv - ∫v du, substitute u = x^2, v = (1/4)sin(4x), and du = 2x dx. This gives: ∫x^2cos(4x)dx = (x^2/4)sin(4x) - ∫(1/4)(2x)sin(4x)dx.

Step 4: Simplify the remaining integral. The new integral is ∫x sin(4x)dx, which again requires integration by parts. Let u = x and dv = sin(4x)dx. Then, du = dx and v = -cos(4x)/4.

Step 5: Apply integration by parts again. Substitute into the formula ∫u dv = uv - ∫v du for the new integral. Combine all terms and simplify to get the final expression: (x^2/4)sin(4x) + (1/4)cos(4x) - (1/16)sin(4x) + C.

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