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

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

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

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

\frac{2 π + 3 3}{24}

Decimal Notation

0.47830 \dots

Solution steps

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

Apply Trigonometric Substitution

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

= \int_{0}^{\frac{π}{6}} cos^{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}

\int (1 - t) cos (t^{2} - 2 t) d t

\frac{d}{d x} (cos (x π))

\int e^{1.25 x} - 4 x d x

\frac{\partial}{\partial x} (6 x^{\frac{1}{3}} y^{\frac{1}{2}})

\int \frac{e ^{x}}{1 + 25 e ^{2 x}} d x

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