Question

How do you find the antiderivative of `int xsqrt(1-x^2) dx`?

Answer 1

The answer is `=-1/3(1-x^2)^(3/2)+C`

Explanation:

We perform this by substitution

Let `u=1-x^2`

`du=-2xdx`

`xdx=-1/2du`

Therefore,

`intxsqrt(1-x^2)dx=-1/2intsqrtudu`

`=-1/2*u^(3/2)/(3/2)`

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

Answer 2

`-1/3(1 - x^2)^(3/2) + C`

Explanation:

Here's an alternative answer using trig substitution.

Let `x = sintheta` . Then `dx = costhetad theta`.

`=intsinthetasqrt(1 - sin^2theta) * costheta d theta`

`=intsintheta sqrt(cos^2theta) * costheta d theta`

`=int sintheta costhetacostheta d theta`

`=int sin thetacos^2theta d theta`

Now make a substitution. Let `u = costheta`, then `du = -sintheta d theta -> d theta = (du)/(-sintheta)`.

`=int sin theta cos^2theta * (du)/(-sintheta)`

`=-int u^2du`

`=-1/3u^3 + C`

`=-1/3cos^3theta + C`

We know from our initial substitution that `x/1 = sintheta`. This means that the side adjacent `theta` in our imaginary triangle measures `sqrt(1 - x^2)`. Therefore, `costheta = sqrt(1 - x^2)`

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

Hopefully this helps!