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

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

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integral of sin(x)cos(1/2 x)

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

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

Solution steps

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

Use the following identity:

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

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

Simplify

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = (- y)

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

x \to 0 lim (x - 1)

\int \frac{4 x ^{3} - 3 x ^{2} - 8 x}{6 x} d x

\int (sec (3 y) tan (36)) d y

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