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

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

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

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

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

Solution steps

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

Simplify

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int e^{\frac{5}{x}} d x

\frac{1}{x ^{5}}

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