#\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!

Related questions
See all questions in Integration by Trigonometric Substitution
Impact of this question
2020 views around the world
You can reuse this answer
Creative Commons License