What is the integral of 2/3 (1-x^2)^{3/2} ?

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

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integral of 2/3 (1-x^2)^{3/2}

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

\frac{2}{3} x (1 - x^{2})^{\frac{3}{2}} + \frac{1}{16} (4 arcsin (x) - sin (4 arcsin (x))) + C

Solution steps

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

Take the constant out:

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

= \frac{2}{3} \cdot \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})^{\frac{1}{2}} d x

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

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

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

Simplify

\frac{2}{3} (x (1 - x^{2})^{\frac{3}{2}} - (- \frac{3}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x))))) : \frac{2}{3} x (1 - x^{2})^{\frac{3}{2}} + \frac{1}{16} (4 arcsin (x) - sin (4 arcsin (x)))

= \frac{2}{3} x (1 - x^{2})^{\frac{3}{2}} + \frac{1}{16} (4 arcsin (x) - sin (4 arcsin (x)))

Add a constant to the solution

= \frac{2}{3} x (1 - x^{2})^{\frac{3}{2}} + \frac{1}{16} (4 arcsin (x) - sin (4 arcsin (x))) + C

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