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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

Simplify

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

Add a constant to the solution

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

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