What is the integral of sqrt((1-x^2)^3) ?

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

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integral of sqrt((1-x^2)^3)

\int (1 - x^{2})^{3} d x

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

Solution steps

\int (1 - x^{2})^{3} d x

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

assuming

(1 - x^{2}) \geq 0

= \int (1 - x^{2})^{\frac{3}{2}} d x

Apply Integration By Parts

: x (1 - x^{2})^{\frac{3}{2}} - \int - 3 x^{2} 1 - x^{2} d x

= x (1 - x^{2})^{\frac{3}{2}} - \int - 3 x^{2} 1 - x^{2} d x

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

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

Simplify

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

Add a constant to the solution

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

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