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

The integral from-1 to 1 of (1-x^2)^{3/2} is (3pi)/8

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

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

\frac{3 π}{8}

Decimal Notation

1.17809 \dots

Solution steps

\int_{- 1}^{1} (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]_{- 1}^{1}

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

\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))))]_{- 1}^{1}

Simplify

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

Compute the boundaries:

\frac{3 π}{8}

= \frac{3 π}{8}

\frac{x}{6 x - 4}, at x = 9

\frac{\partial}{\partial x} (2 x^{4} y z^{3})

x \to 0 - lim (\frac{6}{5 x ^{\frac{1}{3}}})

\int (5 x^{2} - 2 x + 3) d x

What is the integral from-1 to 1 of (1-x^2)^{3/2} ?
The integral from-1 to 1 of (1-x^2)^{3/2} is (3pi)/8