#\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#

computer

1 Answer
Noah G
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!

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