What is the integral of sin(ax)cos(bx) ?

The integral of sin(ax)cos(bx) is 1/2 (-1/(a+b)cos(ax+bx)-1/(a-b)cos(ax-bx))+C

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integral of sin(ax)cos(bx)

\int sin (a x) cos (b x) d x

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

Solution steps

\int sin (a x) cos (b x) d x

Use the following identity:

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

= \int \frac{sin ( a x + b x ) + sin ( a x - b 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 (a x + b x) + sin (a x - b 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 (a x + b x) d x + \int sin (a x - b x) d x)

\int sin (a x + b x) d x = - \frac{1}{a + b} cos (a x + b x)

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

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

Add a constant to the solution

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

What is the integral of sin(ax)cos(bx) ?
The integral of sin(ax)cos(bx) is 1/2 (-1/(a+b)cos(ax+bx)-1/(a-b)cos(ax-bx))+C