What is the integral of cos(x)cos(kx) ?

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

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integral of cos(x)cos(kx)

\int cos (x) cos (k x) d x

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

Solution steps

\int cos (x) cos (k x) d x

Rewrite using trig identities

= \int \frac{cos ( x + k x ) + cos ( x - k 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 cos (x + k x) + cos (x - k 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 cos (x + k x) d x + \int cos (x - k x) d x)

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

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

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

Add a constant to the solution

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

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