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

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

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

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

= 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)))

= 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

f (x) = ln (6 x^{2} + 1)

e^{5 t} t

\frac{d}{d x} (x^{2} - 6 x + 9)

\int_{1}^{2} \frac{1}{x ( ln ( x ) ) ^{1}} d x

\int_{1}^{\infty} \frac{1}{1 + e ^{x}} d x

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