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

Question

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

∫ arcsin(y) dy

Accepted Answer

Welcome back everyone. In this problem, we want to evaluate the integral of the inverse cosine of X with respect to X using integration by parts. Now what do we know about integration by parts? Well, recoil. It basically says that the integral of UDV. Is equal to uV minus the integral of VDU. So if we can assign one of these terms to be you and the other to be DV, then we should be able to substitute to figure out our integration or integral using integration by parts. Now, to solve this integral, we can first identify the parts of the integral and using the lipid rule, and basically it's just a rule that tells us which type of function to prioritize or to choose as you. No, L stands for logarithmic, I inverse trigonometric, then polynomial, then exponential, then trigonometric. Notice here, the integram contains an inverse trigonometric function. So we can treat the integrand as the cosine of inverse x, sorry, the inverse cosine of x multiplied by 1, where 1 is considered a polynomial function, and since inverse trigonometric functions have higher priority in the lipid hierarchy, then we can set u to be equal to the inverse cosine of x. And dV to be equal to 1 multiplied by dx. Now if u equals the inverse cosine of x, then the derivative of u with respect to x would be equal to -1 divided by the square root of 1 minus x2. I know if we multiply both sides by dx, that means DU. Would be equal to, let me just write it here, du would be equal to 1 minus 1 divided by 1 minus the square root of 1 minus x2 multiplied by dx. On the other hand, if DV equals DX, if we integrate both sides, that means V would be equal to x. So now, when we go ahead and apply our integration by parts, then that means the integral of the inverse cosine of x with respect to x. Would be equal to uv. OK, that would be x multiplied by the inverse cosine of x minus the integral of vDu. So that's the integral of x multiplied by the negative value of 1 divided by the square root of 1 minus x 2 with respect to x. And when we simplify that, that's x multiplied by the inverse cosine of x plus the integral of x divided by the square root of 1 minus x2 with respect to x. And now we can try to find this integral using substitution. No solving for this integral, so let me put this in red, actually. OK. So we want to figure out what this integral is. So solving. For the integral of X. Divided by the square root of 1 minus x2 with respect to x. Let's set w to be 1 minus X2. Then That means if we differentiate W with respect to X. OK, then. DWDX equals -2 x. And if that's the case, if we're trying to find DW in terms of dx, then we could rewrite it as negative half of DW equals XDX. So now we have all we need to substitute, so that means our integral of x divided by the square root of 1 minus x2 with respect to x would be equivalent to 12 of the integral of 1 divided by root w with respect tow. We could rewrite this as -52 multiplied by the integral of w to -522 dw, and when we integrate that, we end up with 52 multiplied by w 52 divided by 1/22, which would leave us with w to 1/2 or a negative root w. And since we know W equals 1 minus X2, that means this would be negative root 1 minus X2 plus some constant C. Now that we have this, OK, then we can substitute it back into our original. Integral that we were trying to do by parts. So in other words, that means our integral of the inverse cosine of x with respect to x. Equals X class inverse X. OK. minus the square root of 1 minus x2 + c. So this is going to be the integral of the inverse cosine of X with respect to x. Thanks a lot for watching everyone. I hope this video helped.

Evaluate the integrals in Exercises 1–24 using integration by parts.∫ arcsin(y) dy

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Evaluate the integrals in Exercises 1–24 using integration by parts.
∫ arcsin(y) dy