What is the integral from 0 to 1 of x(1-x^4)^{3/2} ?

The integral from 0 to 1 of x(1-x^4)^{3/2} is (3pi)/(32)

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integral from 0 to 1 of x(1-x^4)^{3/2}

\int_{0}^{1} x (1 - x^{4})^{\frac{3}{2}} d x

\frac{3 π}{32}

Decimal

0.29452 \dots

Solution steps

\int_{0}^{1} x (1 - x^{4})^{\frac{3}{2}} d x

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} (1 - u^{2})^{\frac{3}{2}} d u

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

\frac{3 π}{16}

= \frac{1}{2} \cdot \frac{3 π}{16}

Simplify

= \frac{3 π}{32}

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What is the integral from 0 to 1 of x(1-x^4)^{3/2} ?
The integral from 0 to 1 of x(1-x^4)^{3/2} is (3pi)/(32)