What is the integral from 0 to pi/6 of sin^2(x) ?

The integral from 0 to pi/6 of sin^2(x) is (2pi-3sqrt(3))/(24)

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integral from 0 to pi/6 of sin^2(x)

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

\frac{2 π - 3 3}{24}

Decimal

0.04529 \dots

Solution steps

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

Rewrite using trig identities

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

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 x) d x

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 x - \int_{0}^{\frac{π}{6}} cos (2 x) d x)

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

\int_{0}^{\frac{π}{6}} cos (2 x) d x = \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}

x \to 0 lim (\frac{5 ^{x} - 1 0 ^{x}}{x})

x \to 5 lim (x^{3} + 2 x)

\int \frac{1}{2 ( x ^{3} + x ^{2} )} d x

y^{\frac{1}{2}} \frac{d y}{d x} + y^{\frac{3}{2}} = 1, y (0) = 9

17 e^{x} - e x^{e}

What is the integral from 0 to pi/6 of sin^2(x) ?
The integral from 0 to pi/6 of sin^2(x) is (2pi-3sqrt(3))/(24)