What is the integral of sin(x/4)cos((3x)/4) ?

The integral of sin(x/4)cos((3x)/4) is -1/2 cos(x)+cos(x/2)+C

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integral of sin(x/4)cos((3x)/4)

\int sin (\frac{x}{4}) cos (\frac{3 x}{4}) d x

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

Solution steps

\int sin (\frac{x}{4}) cos (\frac{3 x}{4}) d x

Use the following identity:

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

= \int \frac{sin ( \frac{x}{4} + \frac{3 x}{4} ) + sin ( \frac{x}{4} - \frac{3 x}{4} )}{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}{4} + \frac{3 x}{4}) + sin (\frac{x}{4} - \frac{3 x}{4}) d x

Simplify

sin (\frac{x}{4} + \frac{3 x}{4}) + sin (\frac{x}{4} - \frac{3 x}{4}) : sin (x) - sin (\frac{x}{2})

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of sin(x/4)cos((3x)/4) ?
The integral of sin(x/4)cos((3x)/4) is -1/2 cos(x)+cos(x/2)+C