Question

`\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`

Answer 1

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!