What is the integral from 0 to 1 of z^4sqrt(1-z^2) ?

The integral from 0 to 1 of z^4sqrt(1-z^2) is pi/(32)

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integral from 0 to 1 of z^4sqrt(1-z^2)

\int_{0}^{1} z^{4} 1 - z^{2} d z

\frac{π}{32}

Decimal

0.09817 \dots

Solution steps

\int_{0}^{1} z^{4} 1 - z^{2} d z

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Expand

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

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

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

\int_{0}^{\frac{π}{2}} sin^{6} (u) d u = \frac{5 π}{32}

= \frac{3 π}{16} - \frac{5 π}{32}

Simplify

\frac{3 π}{16} - \frac{5 π}{32} : \frac{π}{32}

= \frac{π}{32}

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\int_{0}^{0.4} x^{2} d x

\int_{0}^{1} 7 x^{3} d x

What is the integral from 0 to 1 of z^4sqrt(1-z^2) ?
The integral from 0 to 1 of z^4sqrt(1-z^2) is pi/(32)