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

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

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integral of sin(x/3)cos((2x)/3)

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

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

Solution steps

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

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

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

Simplify

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

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

\int sin (x) d x = - cos (x)

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

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

Simplify

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

Add a constant to the solution

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

\int x (e^{x}) d x

f (x) = \frac{2}{x - 1}, at x = - 2

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

x \to 0 lim (1 + sin^{2} (3 x))

\frac{d ^{2}}{d x ^{2}} (sin (x) cos (x))

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