Evaluate the integrals in Exercises 1–6.
∫ x arcsin x dx...

Question

Evaluate the integrals in Exercises 1–6.

∫ x arcsin x dx

Accepted Answer

Welcome back everyone. In this problem we want to evaluate the integral of x multiplied by the inverse tangent of X with respect to X. Now for us to evaluate this integral, it would be best to use integration by parts. What do we know? Well, recall that in integration by parts we are, we know that the integral of UDV will be equal to UV minus the integral of VDU. So if we can assign you and DV to each of these terms, then we should be able to integrate by parts and solve. Now which one should be you and which one should be DV? Well, we can follow the LIATE rule. In order of importance or priority, we would set u to be logarithms. Next, inverse trig inverse trig function next, an algebraic term next, a trigonometric term, and finally, an exponential term. Now if we note here, we have X, which is an algebraic term, and the inverse tangent of X, an inverse trigonometric term. So this is what we would set as you. So let's let you. Be equal to the inverse tangent of x. And, in that case, we set dV to be equal to XDX. Now in this case if u equals the inverse tangent of x, then the derivative of u with respect to x will be equal to 1 divided by 1 + x2, which means that duu then will be equal to 1 divided by 1 + x2 dx. On the other hand, if DV equals XDX, then if we integrate both sides. That means V will be equal to 1/2 of x2 because 1/2 of x2 is the integral of x with respect to x. So now that we have D U and V, then we can apply integration by parts because no, this means that the integral of x multiplied by the inverse tangent of x with respect to x. Will be equal to UV, that's the inverse tangent of x. OK. So it's the inverse tangent of X. Multiplied by 1/2 of X2, OK. Minus the integral of vDu minus the integral of 12 of x2 multiplied by 1 divided by 1 + x2 with respect to x. And I know if we simplify that, then we'll have a 12 of x 2 multiplied by the inverse tangent of x. Minus a half of the integral of x2 divided by 1 + x2 with respect to x. So all we need to do to solve no is to figure out the integral of x2 divided by 1 + x2 with respect to x. How can we do that? Well, let's use a common algebraic trick to add and subtract 1 in the numerator to simplify x2 divided by 1 plus. X2, OK, now, if we take it, so that's 1. Sorry, X2. Divided by 1 + X2. And we add and subtract 1 in the numerator that will be X2 + 1 minus 1, OK. Divided by 1 + X2. Now, using what we know about fractions, this will be x2 + 1 divided by 1 + x2 or X2 + 1 minus 1 divided by 1 + X2, and in this case, our first term will simplify to 1, so we'll end up with 1 minus 1 + X2. So no, really, that means the integral of X2 divided by 1 + X2 with respect to X is going to be the integral of 1 minus 1 divided by 1 + X2 with respect to x. And know when we integrate the integral of 1 is x and the integral of 1 divided by 1 + x 2 will be the inverse tangent of x because remember the derivative of the inverse tangent of x would be 1 divided by 1 + x 2. Now that we have this, we can plug it back into our main equation. Which means then that the integral of x multiplied by the inverse tangent of x with respect to x will be equal to 12 of x 2 multiplied by the inverse tangent of x. Minus 122 multiplied by x minus the inverse tangent of x plus c. If we distribute our half, OK. Then we'll end up with minus 1/2 of x, OK, plus 122. Of the inverse tangent of X. Right, and now we can factor out a half from our entire term to end up with a 1/2 multiplied by x2 multiplied by the inverse tangent of x minus x plus the inverse tangent of x. Lai This is going to be the integral of x multiplied by the inverse tangent of x with respect to x. Thanks a lot for watching everyone. I hope this video helped.

Evaluate the integrals in Exercises 1–6.∫ x arcsin x dx

Watch next

Evaluate the integrals in Exercises 1–6.
∫ x arcsin x dx