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

The integral of (sqrt(4-x^2))^3 is x(4-x^2)^{3/2}+6(arcsin(1/2 x)-1/4 sin(4arcsin(1/2 x)))+C

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

\int (4 - x^{2})^{3} d x

x (4 - x^{2})^{\frac{3}{2}} + 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x))) + C

Solution steps

\int (4 - x^{2})^{3} d x

Apply Integration By Parts

: x (4 - x^{2})^{3} - \int - 3 x^{2} 4 - x^{2} d x

= x (4 - x^{2})^{3} - \int - 3 x^{2} 4 - x^{2} d x

\int - 3 x^{2} 4 - x^{2} d x = - 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x)))

= x (4 - x^{2})^{3} - (- 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x))))

Simplify

x (4 - x^{2})^{3} - (- 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x)))) : x (4 - x^{2})^{\frac{3}{2}} + 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x)))

= x (4 - x^{2})^{\frac{3}{2}} + 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x)))

Add a constant to the solution

= x (4 - x^{2})^{\frac{3}{2}} + 6 (arcsin (\frac{1}{2} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} x))) + C

\frac{d}{d x} (\frac{( 4 x ^{2} + 4 x + 4 )}{x})

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

\int_{0}^{\frac{1}{2}} - 2 arccos (x) d x

\int_{0}^{\infty} 9 e^{- 9 x} d x

\frac{\partial}{\partial x} (1 - \frac{3}{y} + x)

What is the integral of (sqrt(4-x^2))^3 ?
The integral of (sqrt(4-x^2))^3 is x(4-x^2)^{3/2}+6(arcsin(1/2 x)-1/4 sin(4arcsin(1/2 x)))+C