Question

How do you find the antiderivative of `int xsqrt(100-x^2)dx`?

Answer 1

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

Explanation:

`I=intxsqrt(100-x^2)dx`

The best substitution to make here is `u=100-x^2`. This implies that `(du)/dx=-2x`, so `du=-2xdx`.

Since we have just `xdx` in the integral, multiply the integral's interior by `-2`. To balance this out, multiply the exterior of the integral by `-1//2`.

`I=-1/2int(-2x)sqrt(100-x^2)dx`

`I=-1/2intunderbrace(sqrt(100-x^2))_sqrtuoverbrace((-2xdx))^(du)`

`I=-1/2intsqrtudu`

This can be integrated if we write the square root using a fractional power.

`I=-1/2intu^(1/2)du`

This can be integrated using the power rule for integration, or `intu^ndu=u^(n+1)/(n+1)+C`. So:

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

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

Since `u=100-x^2`:

`I=-(100-x^2)^(3/2)/3+C`