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

The integral from-1 to 1 of x^4sqrt(1-x^2) is pi/(16)

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

\int_{- 1}^{1} x^{4} 1 - x^{2} d x

\frac{π}{16}

Decimal

0.19634 \dots

Solution steps

\int_{- 1}^{1} x^{4} 1 - x^{2} d x

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

\int_{- \frac{π}{2}}^{\frac{π}{2}} sin^{4} (u) d u = \frac{3 π}{8}

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

= \frac{3 π}{8} - \frac{5 π}{16}

Simplify

\frac{3 π}{8} - \frac{5 π}{16} : \frac{π}{16}

= \frac{π}{16}

\int_{- 4}^{5} - x^{2} + x + 20 d x

\int_{4}^{6} x^{2} + 2 x - 2 d x

\int_{0}^{2} x^{4} - 8 x d x

\int_{0}^{1} 4 + 3 x^{2} d x

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

What is the integral from-1 to 1 of x^4sqrt(1-x^2) ?
The integral from-1 to 1 of x^4sqrt(1-x^2) is pi/(16)