What is the integral of sin(1/4 x)cos(3/4 x) ?

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

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integral of sin(1/4 x)cos(3/4 x)

\int sin (\frac{1}{4} x) cos (\frac{3}{4} x) d x

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

Solution steps

\int sin (\frac{1}{4} x) cos (\frac{3}{4} x) d x

Use the following identity:

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

= \int \frac{sin ( \frac{1}{4} x + \frac{3}{4} x ) + sin ( \frac{1}{4} x - \frac{3}{4} 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{1}{4} x + \frac{3}{4} x) + sin (\frac{1}{4} x - \frac{3}{4} x) d x

Simplify

sin (\frac{1}{4} x + \frac{3}{4} x) + sin (\frac{1}{4} x - \frac{3}{4} x) : sin (x) - sin (\frac{1}{2} x)

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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