What is the integral of sin^2(1/3 x) ?

The integral of sin^2(1/3 x) is 1/2 (x-3/2 sin((2x)/3))+C

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integral of sin^2(1/3 x)

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

\frac{1}{2} (x - \frac{3}{2} sin (\frac{2 x}{3})) + C

Solution steps

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

Rewrite using trig identities

= \int \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 1 - cos (\frac{2}{3} x) d x

Rewrite using trig identities

= \frac{1}{2} \cdot \int 2 sin^{2} (\frac{x}{3}) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot 2 \cdot \int sin^{2} (\frac{x}{3}) d x

Rewrite using trig identities

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{2} (1 - cos (\frac{2 x}{3})) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot 2 \cdot \frac{1}{2} \cdot \int 1 - cos (\frac{2 x}{3}) 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} \cdot 2 \cdot \frac{1}{2} (\int 1 d x - \int cos (\frac{2 x}{3}) d x)

\int 1 d x = x

\int cos (\frac{2 x}{3}) d x = \frac{3}{2} sin (\frac{2 x}{3})

= \frac{1}{2} \cdot 2 \cdot \frac{1}{2} (x - \frac{3}{2} sin (\frac{2 x}{3}))

Simplify

\frac{1}{2} \cdot 2 \cdot \frac{1}{2} (x - \frac{3}{2} sin (\frac{2 x}{3})) : \frac{1}{2} (x - \frac{3}{2} sin (\frac{2 x}{3}))

= \frac{1}{2} (x - \frac{3}{2} sin (\frac{2 x}{3}))

Add a constant to the solution

= \frac{1}{2} (x - \frac{3}{2} sin (\frac{2 x}{3})) + C

\frac{s}{( s ^{2} + 5 ^{2} ) ^{2}}

\int \frac{4 + x ^{2}}{x} d x

\frac{d}{d x} (\frac{x}{( x - 1 )})

x \to 0 lim (\frac{4 e ^{x} - 4}{x})

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

What is the integral of sin^2(1/3 x) ?
The integral of sin^2(1/3 x) is 1/2 (x-3/2 sin((2x)/3))+C