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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int 2 (sin (3 u) - sin (u)) d u

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

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 (3 u) - sin (u) d u

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 (\int sin (3 u) d u - \int sin (u) d u)

\int sin (3 u) d u = - \frac{1}{3} cos (3 u)

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

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

Substitute back

u = \frac{x}{2}

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

Simplify

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

\frac{1}{x ^{2}}, 3

t^{2} y^{^{''}} + 4 t y^{^{'}} - 18 y = 0

\frac{\partial}{\partial x} (- 2 x e^{- 3 x y})

\int 2 x (x^{2} + 15)^{7} d x

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