What is the integral from 0 to pi/(12) of sin^2(u) ?

The integral from 0 to pi/(12) of sin^2(u) is (pi-3)/(24)

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integral from 0 to pi/(12) of sin^2(u)

\int_{0}^{\frac{π}{12}} sin^{2} (u) d u

\frac{π - 3}{24}

Decimal Notation

0.00589 \dots

Solution steps

\int_{0}^{\frac{π}{12}} sin^{2} (u) d u

Rewrite using trig identities

= \int_{0}^{\frac{π}{12}} \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}{12}} 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}{12}} 1 d u - \int_{0}^{\frac{π}{12}} cos (2 u) d u)

\int_{0}^{\frac{π}{12}} 1 d u = \frac{π}{12}

\int_{0}^{\frac{π}{12}} cos (2 u) d u = \frac{1}{4}

= \frac{1}{2} (\frac{π}{12} - \frac{1}{4})

Simplify

\frac{1}{2} (\frac{π}{12} - \frac{1}{4}) : \frac{π - 3}{24}

= \frac{π - 3}{24}

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What is the integral from 0 to pi/(12) of sin^2(u) ?
The integral from 0 to pi/(12) of sin^2(u) is (pi-3)/(24)