\intx^3\sqrt(1-x^2)dx?

I tried to get the answer found on Symbolab, but instead I got:
-1/3(1-x^2)^(3/2)+1/5(1-x^2)^(5/2)+C

computercomputer

1 Answer
Apr 13, 2018

Your answer is excellent and correct!

Explanation:

You're going to have to use trig substitution for this question.
Let x= sintheta. Then dx =costheta d theta.

I = int sin^3theta sqrt(1 - sin^2theta) costheta d theta

I = int sin^3theta sqrt(cos^2theta)costheta d theta

I = int sin^3theta cos^2thetad theta

I = int sintheta(1 -cos^2theta)cos^2thetad theta

I = int (sin theta - sinthetacos^2theta)cos^2thetad theta

I = int sinthetacos^2theta - sinthetacos^4thetad theta

I= int sin thetacos^2theta d theta - int sinthetacos^4theta d theta

Let u = costheta. You should be left with du = -sintheta d theta and d theta = (du)/(-sintheta).

I = -int u^2 du +int u^4 du

I = -1/3u^3 + 1/5u^5 + C

I = -1/3cos^3theta + 1/5cos^5theta + C

Recall from our initial substitution that x= sintheta, and since cos^2x+ sin^2x = 1, we can see that costheta = sqrt(1 - x^2).

I =1/5(1 - x^2)^(5/2) -1/3(1- x^2)^(3/2) + C

I checked and our answer is the same as the one shown on symbolab, except ours is simplified a little further.

Hopefully this helps!