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

The integral from 0 to 1/2 of (x^2)/(sqrt(1-x^2)) is (2pi-3sqrt(3))/(24)

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integral from 0 to 1/2 of (x^2)/(sqrt(1-x^2))

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

\frac{2 π - 3 3}{24}

Decimal

0.04529 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} sin^{2} (u) d u

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

Rewrite using trig identities

= \int_{0}^{\frac{π}{6}} \frac{1 - cos ( 2 u )}{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}^{\frac{π}{6}} 1 - cos (2 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

= \frac{1}{2} (\int_{0}^{\frac{π}{6}} 1 d u - \int_{0}^{\frac{π}{6}} cos (2 u) d u)

\int_{0}^{\frac{π}{6}} 1 d u = \frac{π}{6}

\int_{0}^{\frac{π}{6}} cos (2 u) d u = \frac{3}{4}

= \frac{1}{2} (\frac{π}{6} - \frac{3}{4})

Simplify

\frac{1}{2} (\frac{π}{6} - \frac{3}{4}) : \frac{2 π - 3 3}{24}

= \frac{2 π - 3 3}{24}

\frac{d y}{d x} = 3 y - y^{2}

\int \frac{ln ( 7 x )}{x} d x

y = cos (x)

\frac{d}{d t} (- 3 t)

f (x) = \frac{x}{5}

What is the integral from 0 to 1/2 of (x^2)/(sqrt(1-x^2)) ?
The integral from 0 to 1/2 of (x^2)/(sqrt(1-x^2)) is (2pi-3sqrt(3))/(24)