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

The integral of sin(x/2)cos((2x)/3) is 1/2 (-6/7 cos((7x)/6)+6cos(x/6))+C

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

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

\frac{1}{2} (- \frac{6}{7} cos (\frac{7 x}{6}) + 6 cos (\frac{x}{6})) + C

Solution steps

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

Simplify

sin (\frac{x}{2} + \frac{2 x}{3}) + sin (\frac{x}{2} - \frac{2 x}{3}) : sin (\frac{7 x}{6}) - sin (\frac{x}{6})

= \frac{1}{2} \cdot \int sin (\frac{7 x}{6}) - sin (\frac{x}{6}) 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{7 x}{6}) d x - \int sin (\frac{x}{6}) d x)

\int sin (\frac{7 x}{6}) d x = - \frac{6}{7} cos (\frac{7 x}{6})

\int sin (\frac{x}{6}) d x = - 6 cos (\frac{x}{6})

= \frac{1}{2} (- \frac{6}{7} cos (\frac{7 x}{6}) - (- 6 cos (\frac{x}{6})))

Simplify

= \frac{1}{2} (- \frac{6}{7} cos (\frac{7 x}{6}) + 6 cos (\frac{x}{6}))

Add a constant to the solution

= \frac{1}{2} (- \frac{6}{7} cos (\frac{7 x}{6}) + 6 cos (\frac{x}{6})) + C

81 m^{8} + 2 m^{4} + 1

arcsin (4 x + 1)

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

\int (7 + x^{5})^{6} 5 x^{4} d x

\frac{d}{d x} (cos (\frac{1}{2} x))

What is the integral of sin(x/2)cos((2x)/3) ?
The integral of sin(x/2)cos((2x)/3) is 1/2 (-6/7 cos((7x)/6)+6cos(x/6))+C