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Problem 24 Easy Difficulty

Use the Table of Integrals on Reference Pages 6-10 to evaluate the integral.

$ \displaystyle \int x^3 \arcsin (x^2)\ dx $


$\frac{2 x^{4}-1}{8} \arcsin x^{2}+\frac{x^{2} \sqrt{1-x^{4}}}{8}+C$


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Video Transcript

Okay, so this question wants us to compute this integral. So what we need to dio is find a substitution that turns this into one that we can find its value easily from our table. So let's pick you to be the arguments of the arc sine so X squared. So if we take to you, we get two x d x, so 1/2 d'you is equivalent to x t X. So then let's write this integral in a slightly different form so we can see what happens here. So we see that X squared is you and X d X is 1/2 to you. So this becomes 1/2 times the integral of you Sign in verse of you, do you? So now from there we can use the following formula from our table. So this tells us what the integral of a signed inverse times you is. So so are integral is 1/2 times this thing. So now all we have to do is distribute the two to both terms and replace you with X squared. So this gives us two X to the fourth because you squared squared. We're sorry. Ex word squares except the fourth minus one over a sine inverse of X squared plus X squared square root one minus X to the fourth all over Hey plus C, and that's our final answer.

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