What is the integral of x^3sqrt(16-4x^2) ?

The integral of x^3sqrt(16-4x^2) is -2(-1/5 (-x^2+4)^{5/2}+4/3 (-x^2+4)^{3/2})+C

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integral of x^3sqrt(16-4x^2)

\int x^{3} 16 - 4 x^{2} d x

- 2 (- \frac{1}{5} (- x^{2} + 4)^{\frac{5}{2}} + \frac{4}{3} (- x^{2} + 4)^{\frac{3}{2}}) + C

Solution steps

\int x^{3} 16 - 4 x^{2} d x

Apply u-substitution

= \int \frac{2 u ^{\frac{1}{3}} - u ^{\frac{2}{3}} + 4}{3} d u

Take the constant out:

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

= \frac{2}{3} \cdot \int u^{\frac{1}{3}} - u^{\frac{2}{3}} + 4 d u

Apply u-substitution

= \frac{2}{3} \cdot \int 3 v^{3} - v^{2} + 4 d v

Take the constant out:

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

= \frac{2}{3} \cdot 3 \cdot \int v^{3} - v^{2} + 4 d v

Apply u-substitution

= \frac{2}{3} \cdot 3 \cdot \int - (- w^{2} + 4) w^{2} d w

Take the constant out:

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

= \frac{2}{3} \cdot 3 (- \int (- w^{2} + 4) w^{2} d w)

Expand

(- w^{2} + 4) w^{2} : - w^{4} + 4 w^{2}

= \frac{2}{3} \cdot 3 (- \int - w^{4} + 4 w^{2} d w)

Apply the Sum Rule:

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

= \frac{2}{3} \cdot 3 (- (- \int w^{4} d w + \int 4 w^{2} d w))

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

\int 4 w^{2} d w = \frac{4 w ^{3}}{3}

= \frac{2}{3} \cdot 3 (- (- \frac{w ^{5}}{5} + \frac{4 w ^{3}}{3}))

Substitute back

= \frac{2}{3} \cdot 3 - - \frac{( - ( ( x ^{3} ) ^{\frac{1}{3}} ) ^{2} + 4 ) ^{5}}{5} + \frac{4 ( - ( ( x ^{3} ) ^{\frac{1}{3}} ) ^{2} + 4 ) ^{3}}{3}

Simplify

\frac{2}{3} \cdot 3 - - \frac{( - ( ( x ^{3} ) ^{\frac{1}{3}} ) ^{2} + 4 ) ^{5}}{5} + \frac{4 ( - ( ( x ^{3} ) ^{\frac{1}{3}} ) ^{2} + 4 ) ^{3}}{3} : - 2 (- \frac{1}{5} (- x^{2} + 4)^{\frac{5}{2}} + \frac{4}{3} (- x^{2} + 4)^{\frac{3}{2}})

= - 2 (- \frac{1}{5} (- x^{2} + 4)^{\frac{5}{2}} + \frac{4}{3} (- x^{2} + 4)^{\frac{3}{2}})

Add a constant to the solution

= - 2 (- \frac{1}{5} (- x^{2} + 4)^{\frac{5}{2}} + \frac{4}{3} (- x^{2} + 4)^{\frac{3}{2}}) + C

What is the integral of x^3sqrt(16-4x^2) ?
The integral of x^3sqrt(16-4x^2) is -2(-1/5 (-x^2+4)^{5/2}+4/3 (-x^2+4)^{3/2})+C