Evaluate the integrals in Exercises 1–24 using integration by ...

Question

Evaluate the integrals in Exercises 1–24 using integration by parts.

∫ x² sin(x) dx

Accepted Answer

Welcome back everyone. In this problem, we want to evaluate the integral of X2 + X with respect to X using integration by parts. Now what do we know about integration by parts that will help us? Well, recall that the integration by parts formula tells us that the integral of UDV. Is equal to UV minus the integral of VDU. So if we can set one of these terms to be you and the other to be DV, then we should be able to substitute into our formula for integration by parts to solve. How can we do that? Well, let's first identify the parts of the integrand using lipid. In other words, logs, inverse, trig, polynomial, exponential, and trigonometric rule, and this rule helps us to prioritize which one of the terms we would first select as you. Now notice here that our integran consists of a polynomial x 2 and a trigonometric function x. Since the polynomial comes before the trigonometric function in our little acronym or rule here, then it would be best to set u as x 2, the polynomial. And thus set V or sorry, DV. As Cos XDX. I know we're going to go ahead and figure it out, but bear in mind that because the polynomial is of the 2nd degree, we will need to apply the integration by parts formula a 2nd time to the resulting integral to reach the final solution. So let's apply it here for the 1st time. Now if U equals x 2, that means the derivative of u with respect to x equals 2 x, so duu would be equal to 2 x dx. On the other hand, if dV equals cus x dx, if we integrate both sides, then we should get v to be equal to sin x. So now, if we apply the integration by parts formula, that means the integral of x 2 cost x with respect to x. Equals UV, so that's x2 multiplied by sin x minus the integral of VDU sinx multiplied by 2 x dx. And now we could rewrite this as x2 sin x minus 2 multiplied by the integral of x sinx with respect to x. So this is the point where we'll need to integrate this term again by parts. Now we follow the same procedure, right, uh, to do this, OK, so let me put this down here, so integrating. X X. With respect to X by parts. Again, we choose our polynomial to be you, so let u equal x and. Or trig function to be DV, so DV equals sin xDX. Now if U equals x, that means the derivative of u with respect to x is 1, so DU equals dx, and if dV equals the sin x dx, then when we integrate that, we should get V to be equal to negative cost x. So no applying our integration by parts formula that means the integral of x sinx with respect to x would be equal to x multiplied by a negative cost xuV minus the integral of VDU, so that's negative cost xDX. And now if we think about it, that would be g x x plus the integral of cost xDX. Which equals x x plus sin x. So now when we substitute this second integration by parts into our first expression here, our original expression. Then this means that the integral of x2 cost xDX. Is going to be x 2 sin x minus 2 multiplied by g x x plus sin x, OK. Plus C And when we simplify, that's going to be x grade sin x. + 2 x x minus 2 sin x + c. This is our integral of x2 cost X with respect to x using integration by parts. Thanks a lot for watching everyone. I hope this video helped.

Evaluate the integrals in Exercises 1–24 using integration by parts.∫ x² sin(x) dx

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Evaluate the integrals in Exercises 1–24 using integration by parts.
∫ x² sin(x) dx