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

The integral of sin(y)cos(y/2) is -4/3 cos^3(y/2)+C

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integral of sin(y)cos(y/2)

\int sin (y) cos (\frac{y}{2}) d y

- \frac{4}{3} cos^{3} (\frac{y}{2}) + C

Solution steps

\int sin (y) cos (\frac{y}{2}) d y

Apply u-substitution

: \int sin (2 u) cos (u) \cdot 2 d u

= \int sin (2 u) cos (u) \cdot 2 d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int sin (2 u) cos (u) d u

Rewrite using trig identities

= 2 \cdot \int 2 cos^{2} (u) sin (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot 2 \cdot \int cos^{2} (u) sin (u) d u

Apply u-substitution

: \int - v^{2} d v

= 2 \cdot 2 \cdot \int - v^{2} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot 2 (- \int v^{2} d v)

Apply the Power Rule

: \frac{v ^{3}}{3}

= 2 \cdot 2 (- \frac{v ^{3}}{3})

Substitute back

= 2 \cdot 2 (- \frac{cos ^{3} ( \frac{y}{2} )}{3})

Simplify

2 \cdot 2 (- \frac{cos ^{3} ( \frac{y}{2} )}{3}) : - \frac{4}{3} cos^{3} (\frac{y}{2})

= - \frac{4}{3} cos^{3} (\frac{y}{2})

Add a constant to the solution

= - \frac{4}{3} cos^{3} (\frac{y}{2}) + C

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