Evaluate the integrals in Exercises 1–8 using integration by p...

Question

Evaluate the integrals in Exercises 1–8 using integration by parts.

∫ x sin(x) cos(x) dx

Accepted Answer

Welcome back everyone. In this problem, we want to evaluate the integral of x sin x + 2 x with respect to x using integration by parts. Now before we use the standard integration by parts formula to evaluate this integral, let's first try to simplify this trigonometric part to make it more manageable. Now what do we know about the sign of something being multiplied by the cosine of the other that we could rewrite it in a simpler way? Well recall the product to some identity that tells us that the sign of a. Multiplied by the cosine of b. Is equal to a half of the sin of a plus b. Plus the sin of a minus b. So we can apply that here to rewrite the product of sinx and the cosine of 2 x in terms of sine because no, this means then. That if we view a as x and b as 2 x, then the sin of x multiplied by the cosine of 2 x will be equal to a half of the sin of x plus 2x or 3 x plus the sin of x minus 2x or negative x. No, since the sin of x is equal to the negative sin of x, we could take this a step further and rewrite this as a half of the sin of 3 x minus the sin of x. And if we apply that to our integral, OK. Then the integral of x sinx costs 2 x with respect to x. Could be rewritten. OK, as the integral of x multiplied by 12, OK, of the sin of 3 x minus the sin of x, OK. With respect to X. And now we can split that into two separate terms, that is. The half of the integral of x sin of 3 x with respect to x, OK, minus a half of the integral of x sinx with respect to x. So if we can integrate both of these, then we can combine them and we should be able to solve. Now we can integrate both of them by parts. So let's go ahead and do that. So integrating both terms. Using integration by parts. And now we can just remind ourselves that when we say integration by parts, what that basically means to integrate using integration by parts, the integral of UDV it will be equal to UV minus the integral of VDU. That's the formula for integration by parts. So here if we can apply this formula to both of these, then we can integrate. So let's start with our first integral, that is, let me put this in red here. Let's start with the integral of X. Multiplied by the sine of 3 x. OK, with respect to x, how can we integrate here? Well, when we use integration by parts for this subinterval, it would be helpful for us to use the LIATE rule, OK, to pick our values of U and DV because basically it's a way for us to strategically choose the best value to use for you and the best value to use for DV. Here L stands for logarithmic, i inverse trigonometric, A algebraic, T trigonometric, and E exponential. Now if we take our term here, the integral of x multiplied by a sin of 3 x with respect to x, notice here x is an algebraic term and the sin of 3 x is a trigonometric term. So based on the order of hierarchy, we would choose an algebraic term over the trigonometric term to set as u. So then let's let you be equal to x. OK, and let's let dV be equal to the sine of 3 x with respect to x, OK, and for this one, let me just put a line here because we're going to separate it and solve our second subintegral on the right hand side. Now, in this case, if u equals x, then the derivative of u with respect to x is 1, so that means duu would be equal to dx. On the other hand, if dV equals the sine of 3 x with respect to x, if we integrate both sides, then we should get V to be equal to -1/3 of the cosine of 3 x, OK. So since we have you and V. And we have DU as well and DV. Then by applying the integration by parts formula, this would be UV. OK. This is going to be equal to UV, which is going to be x multiplied by -1/3 of the cosine of 3 x, OK. Minus the integral of VDU, so that's minus. 1/3 of the cosine of 3 x dx. I know here when we simplify this becomes negative x divided by 3 multiplied by the cosine of 3 x plus 1/3 of the integral of the cosine of 3 x with respect to x. I know when we integrate that here we end up with 1/3 of x multiplied by the cosine of 3 x plus 1/9 of the sin of 3 x, OK, because the integral of the cosine of 3 x with respect to x would be 1/3 of the sine of 3 x. So now we have the integral for that term. Next, let's do the same for our second term, so if we go back here. That is The integral of x sinx with respect to x. Now again we can integrate by parts, so similar, we have an algebraic term coming sorry, and a trigonometric term, so we can set our algebraic term as you. So let U equals x and let's let dV equal the sin of x with respect to x. Now in this case, DU would be equal to dx, sorry, because the derivative of u with respect to x equals 1, similar to what we had before, and no v would be equal to the negative cosine of x. So applying the formula. Then integrating by parts, it will become x multiplied by the negative cosine of x. OK, minus the integral of negative cost X with respect to X. We could rewrite this as negative x class X, plus the integral of class X with respect to X. And when we integrate, we end up with negative x x plus sin x. Now that we have both these terms, we can combine our results. So now, this means then. That. We'll end up with a 12 of. Our first term negative x divided by 3, class of 3 X plus 1/9 of the sin of 3 X, OK. Minus X X, OK, plus sin x. OK. Plus C. So this is our integral, and if we distribute the half and the negative sign, then our result will be g x multiplied by the cosine of 3 x divided by 6. Plus the sine of 3 x divided by 18. Plus X multiplied by the cosine of x divided by 2 minus the sin of x, OK, divided by 2 plus c. This is the integral of x sin x plus 2 x with respect to x. Thanks a lot for watching everyone. I hope this video helped.

Evaluate the integrals in Exercises 1–8 using integration by parts.∫ x sin(x) cos(x) dx

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Evaluate the integrals in Exercises 1–8 using integration by parts.
∫ x sin(x) cos(x) dx