What is the integral from 0 to pi/3 of sin^2(x) ?

The integral from 0 to pi/3 of sin^2(x) is (4pi-3sqrt(3))/(24)

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integral from 0 to pi/3 of sin^2(x)

\int_{0}^{\frac{π}{3}} sin^{2} (x) d x

\frac{4 π - 3 3}{24}

Decimal

0.30709 \dots

Solution steps

\int_{0}^{\frac{π}{3}} sin^{2} (x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{3}} \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}{3}} 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}{3}} 1 d x - \int_{0}^{\frac{π}{3}} cos (2 x) d x)

\int_{0}^{\frac{π}{3}} 1 d x = \frac{π}{3}

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

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

Simplify

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

= \frac{4 π - 3 3}{24}

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

y^{2} - 5 x = 4, x - y = 2

x^{\frac{1}{2}}

\frac{3 x d x}{x ^{2} + 2} = - \frac{y d y}{y ^{2} + 1}

\frac{d}{d x} (\frac{4 x}{x ^{2} + 3})

What is the integral from 0 to pi/3 of sin^2(x) ?
The integral from 0 to pi/3 of sin^2(x) is (4pi-3sqrt(3))/(24)