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

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

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integral of cos(ax)cos(bx)

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

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

Solution steps

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

Rewrite using trig identities

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

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

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

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

Add a constant to the solution

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

x \to 2 lim (\frac{x ^{2} - a}{x - 2})

d er i v a t i v e 2 x^{3} - 3

\frac{\partial}{\partial y} (2 y - 2)

\frac{d}{d x} (e^{x} cos (9 x))

in v erse l a pl a ce \frac{1}{s ^{2} - 4 s + 5}

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