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

2 Answers
Feb 10, 2017

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

Feb 10, 2017

-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!