What is the integral of x^3sqrt(36-x^2) ?

The integral of x^3sqrt(36-x^2) is 1/5 (36-x^2)^{5/2}-12(36-x^2)^{3/2}+C

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integral of x^3sqrt(36-x^2)

\int x^{3} 36 - x^{2} d x

\frac{1}{5} (36 - x^{2})^{\frac{5}{2}} - 12 (36 - x^{2})^{\frac{3}{2}} + C

Solution steps

\int x^{3} 36 - x^{2} d x

Apply u-substitution

: \int - (- u^{2} + 36) u^{2} d u

= \int - (- u^{2} + 36) u^{2} d u

Take the constant out:

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

= - \int (- u^{2} + 36) u^{2} d u

Expand

(- u^{2} + 36) u^{2} : - u^{4} + 36 u^{2}

= - \int - u^{4} + 36 u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int u^{4} d u + \int 36 u^{2} d u)

\int u^{4} d u = \frac{u ^{5}}{5}

\int 36 u^{2} d u = 12 u^{3}

= - (- \frac{u ^{5}}{5} + 12 u^{3})

Substitute back

u = 36 - x^{2}

= - (- \frac{( 36 - x ^{2} ) ^{5}}{5} + 12 (36 - x^{2})^{3})

Simplify

- (- \frac{( 36 - x ^{2} ) ^{5}}{5} + 12 (36 - x^{2})^{3}) : \frac{1}{5} (36 - x^{2})^{\frac{5}{2}} - 12 (36 - x^{2})^{\frac{3}{2}}

= \frac{1}{5} (36 - x^{2})^{\frac{5}{2}} - 12 (36 - x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (36 - x^{2})^{\frac{5}{2}} - 12 (36 - x^{2})^{\frac{3}{2}} + C

x \to 0 lim (\frac{e ^{x} - 1}{13 x})

f (x) = 2 e^{x} cos (x)

\int_{0}^{1} 3 e^{- 4 x} d x

\frac{\partial ^{2}}{\partial x \partial y} (x arcsin (\frac{y}{x}))

\frac{d y}{d t} = e^{4 t + 5 y}

What is the integral of x^3sqrt(36-x^2) ?
The integral of x^3sqrt(36-x^2) is 1/5 (36-x^2)^{5/2}-12(36-x^2)^{3/2}+C