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

The integral from 0 to 1 of x^4(1-x^2)^{3/2} is (3pi)/(256)

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

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

\frac{3 π}{256}

Decimal

0.03681 \dots

Solution steps

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

Apply Trigonometric Substitution

= \int_{0}^{\frac{π}{2}} sin^{4} (u) cos^{4} (u) d u

Rewrite using trig identities

= \int_{0}^{\frac{π}{2}} sin^{4} (u) (1 - sin^{2} (u))^{2} d u

Expand

sin^{4} (u) (1 - sin^{2} (u))^{2} : sin^{4} (u) - 2 sin^{6} (u) + sin^{8} (u)

= \int_{0}^{\frac{π}{2}} sin^{4} (u) - 2 sin^{6} (u) + sin^{8} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{\frac{π}{2}} sin^{4} (u) d u - \int_{0}^{\frac{π}{2}} 2 sin^{6} (u) d u + \int_{0}^{\frac{π}{2}} sin^{8} (u) d u

\int_{0}^{\frac{π}{2}} sin^{4} (u) d u = \frac{3 π}{16}

\int_{0}^{\frac{π}{2}} 2 sin^{6} (u) d u = \frac{5 π}{16}

\int_{0}^{\frac{π}{2}} sin^{8} (u) d u = \frac{35 π}{256}

= \frac{3 π}{16} - \frac{5 π}{16} + \frac{35 π}{256}

Simplify

\frac{3 π}{16} - \frac{5 π}{16} + \frac{35 π}{256} : \frac{3 π}{256}

= \frac{3 π}{256}

What is the integral from 0 to 1 of x^4(1-x^2)^{3/2} ?
The integral from 0 to 1 of x^4(1-x^2)^{3/2} is (3pi)/(256)