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

The integral of sin(x/3)cos(x) is 1/2 (-3/4 cos((4x)/3)+3/2 cos((2x)/3))+C

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

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

\frac{1}{2} (- \frac{3}{4} cos (\frac{4 x}{3}) + \frac{3}{2} cos (\frac{2 x}{3})) + C

Solution steps

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

Use the following identity:

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

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

Simplify

sin (\frac{x}{3} + x) + sin (\frac{x}{3} - x) : sin (\frac{4 x}{3}) - sin (\frac{2 x}{3})

= \frac{1}{2} \cdot \int sin (\frac{4 x}{3}) - sin (\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} (\int sin (\frac{4 x}{3}) d x - \int sin (\frac{2 x}{3}) d x)

\int sin (\frac{4 x}{3}) d x = - \frac{3}{4} cos (\frac{4 x}{3})

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

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

Simplify

= \frac{1}{2} (- \frac{3}{4} cos (\frac{4 x}{3}) + \frac{3}{2} cos (\frac{2 x}{3}))

Add a constant to the solution

= \frac{1}{2} (- \frac{3}{4} cos (\frac{4 x}{3}) + \frac{3}{2} cos (\frac{2 x}{3})) + C

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What is the integral of sin(x/3)cos(x) ?
The integral of sin(x/3)cos(x) is 1/2 (-3/4 cos((4x)/3)+3/2 cos((2x)/3))+C