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

Nov 6, 2016

$- {\left(100 - {x}^{2}\right)}^{\frac{3}{2}} / 3 + C$

#### Explanation:

$I = \int x \sqrt{100 - {x}^{2}} \mathrm{dx}$

The best substitution to make here is $u = 100 - {x}^{2}$. This implies that $\frac{\mathrm{du}}{\mathrm{dx}} = - 2 x$, so $\mathrm{du} = - 2 x \mathrm{dx}$.

Since we have just $x \mathrm{dx}$ in the integral, multiply the integral's interior by $- 2$. To balance this out, multiply the exterior of the integral by $- 1 / 2$.

$I = - \frac{1}{2} \int \left(- 2 x\right) \sqrt{100 - {x}^{2}} \mathrm{dx}$

$I = - \frac{1}{2} \int {\underbrace{\sqrt{100 - {x}^{2}}}}_{\sqrt{u}} {\overbrace{\left(- 2 x \mathrm{dx}\right)}}^{\mathrm{du}}$

$I = - \frac{1}{2} \int \sqrt{u} \mathrm{du}$

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

$I = - \frac{1}{2} \int {u}^{\frac{1}{2}} \mathrm{du}$

This can be integrated using the power rule for integration, or $\int {u}^{n} \mathrm{du} = {u}^{n + 1} / \left(n + 1\right) + C$. So:

$I = - \frac{1}{2} \left({u}^{\frac{1}{2} + 1} / \left(\frac{1}{2} + 1\right)\right) + C = - \frac{1}{2} \left({u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right) + C$

$I = - \frac{1}{2} \left(\frac{2}{3}\right) {u}^{\frac{3}{2}} + C = - {u}^{\frac{3}{2}} / 3 + C$

Since $u = 100 - {x}^{2}$:

$I = - {\left(100 - {x}^{2}\right)}^{\frac{3}{2}} / 3 + C$