79–82. {Use of Tech} Double table look-up The following integr...

Question

79–82. {Use of Tech} Double table look-up The following integrals may require more than one table look-up. Evaluate the integrals using a table of integrals, and then check your answer with a computer algebra system.
79. ∫ x sin⁻¹(2x) dx

79. ∫ x sin⁻¹(2x) dx

Accepted Answer

Welcome back, everyone. In this problem, we want to evaluate the integral of Zarcine 5Z with respect to Z. No, let's first try to integrate by using substitution, OK? So let's let you be equal to the arc sign of 5 Z. Well, in this case, that means z will be equal to 1/5 of sine u, which then means that the derivative of Z with respect to you will be equal to 15th of cou, so Dz will be equal to 1/5 of couu. Now, if we think about it here, we have an expression for Z for the arc sign of 5 Z and for D Z so we can substitute it into our equation. In other words, we could rewrite the the integral of Zarine 5 Z. With respect to Z. As the integral of 1/5 of sine U Z multiplied by Uar sign of 5 Z multiplied by 1/5 of the cosine of UDU or D Z. And if we simplify that, that would be 1/25 of the integral of you sign you cost you. With respect to you. Now we have 3 terms here in our product, right? So let's try to figure out if there's a way to simplify this. If we can recall our trigonometric identities, the product of sine you and cause you are 2 multiplied by the product of sine you and cause you would be equal to the sign of to you, OK? So this means then that sign you costs you. So you cost you. Must be equal to a half of sign to you. So in that case, that must mean then what we could rewrite or integral as 1 25th multiplied by 1/2 or 1/50 of the integral of you sign to you with respect to you. But now we're still multiplying two expressions, two terms. So to simplify, we can integrate by parts. No integrating by parts, OK, integrating by parts. Let's let. A OK, be equal to you. And let's let DB be equal to the sign of to you with respect to you because remember in integration by parts, OK, the integral of ADB would be equal to AB minus the integral of BDA, OK. So in that case, Since we have this for integrating by parts, if A equals U, D A equals DU, and if D B equals sin2 U D U, then B would be equal to the integral of sine 2UDU, which would be equal to a negative half of the cosine of tou. Know that we have that using what we know from integrating by parts, then we can say that 1/50 of the integral of you sign to you with respect to you is going to be equal to 1/50. Of integrating by parts A, B, that is a negative half of U cost to you minus. The integral of BDA, OK. So that would be minus half of the integral of cost to you, OK, with respect to you. No, we still have to integrate here. And when we do that, if we integrate for our second term, so let's write back everything that we already have. OK. But for a second term, this integral is going to be negative 5 multiplied by. Uh, here, or sorry, it would be a half multiplied by 1/2, OK, which is gonna be a 25%, OK. Of the sign of to you. That's after we integrate this. So this is gonna be our result plus C. Now, before we go any further here. You know what, actually, we could expand. So 1/50 multiplied by 1/2, -100 of you cost to you. And 1/4 multiplied by 1/50 is 1200. So that's 1/2/100 of the sign off to you, see. Now if we think about it here, where we, we've, so now we've found our integral in terms of you, all we need to do is to rewrite it in terms of Z. Now, since you, was equal to the arc sign of 5 Z like we said earlier, OK, then that must mean that the sign of 2, which equals 2 sine U costs you, OK. It's going to be equal to. Uh 2 multiplied by 5 z because sine u would be 5 z or 10z multiplied by cost U. No cost u would be the square root of 1 minus sine u squared, so that's the square root of 1 minus 5 z squared or 25 z squared. So now we have an expression for you. We have an expression for assigned to you. Finally, we need an expression for a cost to you. I know if we think about it, cost 2u is going to be equal to 1 minus 2 square u by or trigonometric identities, which would be equal to 1 minus 2 multiplied by 5 z2, which equals 1 minus 50 z squad. So now that we have expressions for cost to you, for assigned to you, and for you, OK. Then if we put them back into our integral, this must mean that our original integral of Zarcine 5 Z with respect to Z. It's going to be equal to -100 multiplied by the arc sign of 5 Z. OK. Multiplied by 1 minus 50 Z. OK, plus 1200. Of 10 Z multiplied by the square root of 1 minus 25 Z squared, OK, plus C. And if we simplify both of our fractions here, both of our quotient, that's going to be 500 minus. 50 Z minus 1 divided by 50 squared minus 1 divided by 100 multiplied by the arcs sign. Uh, 5 Z. Plus Z multiplied by the square root of 1 minus 25 z squared divided by 20 because 10 would go into 200, so 10 would factor out of 200. So we have all of this plus C. This is our integral. Thanks a lot for watching everyone. I hope this video helped.

79–82. {Use of Tech} Double table look-up The following integrals may require more than one table look-up. Evaluate the integrals using a table of integrals, and then check your answer with a computer algebra system.
79. ∫ x sin⁻¹(2x) dx

Key Concepts

Integration by Parts

Inverse Trigonometric Functions

Use of Integral Tables and Computer Algebra Systems

Watch next

79–82. {Use of Tech} Double table look-up The following integrals may require more than one table look-up. Evaluate the integrals using a table of integrals, and then check your answer with a computer algebra system.79. ∫ x sin⁻¹(2x) dx

Key Concepts

Integration by Parts

Inverse Trigonometric Functions

Use of Integral Tables and Computer Algebra Systems

Watch next

79–82. {Use of Tech} Double table look-up The following integrals may require more than one table look-up. Evaluate the integrals using a table of integrals, and then check your answer with a computer algebra system.
79. ∫ x sin⁻¹(2x) dx