What is the integral of sqrt(a^2-r^2)r ?

The integral of sqrt(a^2-r^2)r is -1/3 (a^2-r^2)^{3/2}+C

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integral of sqrt(a^2-r^2)r

\int a^{2} - r^{2} r d r

- \frac{1}{3} (a^{2} - r^{2})^{\frac{3}{2}} + C

Solution steps

\int a^{2} - r^{2} r d r

Apply u-substitution

: \int - \frac{u}{2} d u

= \int - \frac{u}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= - \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = a^{2} - r^{2}

= - \frac{1}{2} \cdot \frac{2}{3} (a^{2} - r^{2})^{\frac{3}{2}}

Simplify

- \frac{1}{2} \cdot \frac{2}{3} (a^{2} - r^{2})^{\frac{3}{2}} : - \frac{1}{3} (a^{2} - r^{2})^{\frac{3}{2}}

= - \frac{1}{3} (a^{2} - r^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{3} (a^{2} - r^{2})^{\frac{3}{2}} + C

\frac{\partial}{\partial x} (10 x + 4 y)

- \frac{4}{x ^{2}}

\int \frac{1}{3 - 7 x} d x

f (x) = arcsin (x^{6})

\frac{\partial}{\partial x} (- \frac{x}{( x ^{2} + y ^{2} + 1 )})

What is the integral of sqrt(a^2-r^2)r ?
The integral of sqrt(a^2-r^2)r is -1/3 (a^2-r^2)^{3/2}+C