What is the integral from 0 to 2pi of sin^2(1/3 x) ?

The integral from 0 to 2pi of sin^2(1/3 x) is (8pi+3sqrt(3))/8

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integral from 0 to 2pi of sin^2(1/3 x)

\int_{0}^{2 π} sin^{2} (\frac{1}{3} x) d x

\frac{8 π + 3 3}{8}

Decimal Notation

3.79111 \dots

Solution steps

\int_{0}^{2 π} sin^{2} (\frac{1}{3} x) d x

Rewrite using trig identities

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

\int_{0}^{2 π} 1 d x = 2 π

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

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

Simplify

\frac{1}{2} (2 π - (- \frac{3 3}{4})) : \frac{8 π + 3 3}{8}

= \frac{8 π + 3 3}{8}

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

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

f (x) = (x - 1) (x^{2} - 8 x + 7)

f (x) = \frac{- cos ( x )}{4 x ^{2} + 2 x - 3}

\frac{1}{( s - 5 ) ^{3}}

What is the integral from 0 to 2pi of sin^2(1/3 x) ?
The integral from 0 to 2pi of sin^2(1/3 x) is (8pi+3sqrt(3))/8