What is the integral of sin(x)*cos(nx) ?

The integral of sin(x)*cos(nx) is 1/2 (-1/(1+n)cos(x+nx)-1/(1-n)cos(x-nx))+C

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integral of sin(x)*cos(nx)

\int sin (x) \cdot cos (n x) d x

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

Solution steps

\int sin (x) cos (n x) d x

Use the following identity:

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

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

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

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

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

Add a constant to the solution

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

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