What is the integral of (sqrt(9-x^2))^3 ?

The integral of (sqrt(9-x^2))^3 is x(9-x^2)^{3/2}+243/8 (arcsin(1/3 x)-1/4 sin(4arcsin(1/3 x)))+C

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integral of (sqrt(9-x^2))^3

\int (9 - x^{2})^{3} d x

x (9 - x^{2})^{\frac{3}{2}} + \frac{243}{8} (arcsin (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{3} x))) + C

Solution steps

\int (9 - x^{2})^{3} d x

Apply Integration By Parts

: x (9 - x^{2})^{3} - \int - 3 x^{2} 9 - x^{2} d x

= x (9 - x^{2})^{3} - \int - 3 x^{2} 9 - x^{2} d x

\int - 3 x^{2} 9 - x^{2} d x = - \frac{243}{8} (arcsin (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{3} x)))

= x (9 - x^{2})^{3} - (- \frac{243}{8} (arcsin (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{3} x))))

Simplify

x (9 - x^{2})^{3} - (- \frac{243}{8} (arcsin (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{3} x)))) : x (9 - x^{2})^{\frac{3}{2}} + \frac{243}{8} (arcsin (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{3} x)))

= x (9 - x^{2})^{\frac{3}{2}} + \frac{243}{8} (arcsin (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{3} x)))

Add a constant to the solution

= x (9 - x^{2})^{\frac{3}{2}} + \frac{243}{8} (arcsin (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{1}{3} x))) + C

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What is the integral of (sqrt(9-x^2))^3 ?
The integral of (sqrt(9-x^2))^3 is x(9-x^2)^{3/2}+243/8 (arcsin(1/3 x)-1/4 sin(4arcsin(1/3 x)))+C