What is the integral from 0 to 1 of (1-x^2)^{3/2} ?

The integral from 0 to 1 of (1-x^2)^{3/2} is (3pi)/(16)

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integral from 0 to 1 of (1-x^2)^{3/2}

\int_{0}^{1} (1 - x^{2})^{\frac{3}{2}} d x

\frac{3 π}{16}

Decimal Notation

0.58904 \dots

Solution steps

\int_{0}^{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]_{0}^{1}

= [x (1 - x^{2})^{\frac{3}{2}} - \int - 3 x^{2} (1 - x^{2})^{\frac{1}{2}} d x]_{0}^{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))))]_{0}^{1}

Simplify

= [x (1 - x^{2})^{\frac{3}{2}} + \frac{3}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x)))]_{0}^{1}

Compute the boundaries:

\frac{3 π}{16}

= \frac{3 π}{16}

\int \frac{15}{x ^{2} + 1} d x

e^{17 x}

f (x) = x ln (x)

\frac{d}{d x} (x e^{2 x^{2} - 4 x})

What is the integral from 0 to 1 of (1-x^2)^{3/2} ?
The integral from 0 to 1 of (1-x^2)^{3/2} is (3pi)/(16)