What is the integral of cos(4x)cos(-3x) ?

The integral of cos(4x)cos(-3x) is 1/2 (sin(x)+1/7 sin(7x))+C

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integral of cos(4x)cos(-3x)

\int cos (4 x) cos (- 3 x) d x

\frac{1}{2} (sin (x) + \frac{1}{7} sin (7 x)) + C

Solution steps

\int cos (4 x) cos (- 3 x) d x

Simplify

cos (4 x) cos (- 3 x) : cos (3 x) cos (4 x)

= \int cos (3 x) cos (4 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (cos (7 x) + cos (- x)) 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 (7 x) + cos (- x) d x

Simplify

cos (7 x) + cos (- x) : cos (x) + cos (7 x)

= \frac{1}{2} \cdot \int cos (x) + cos (7 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) d x + \int cos (7 x) d x)

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

\int cos (7 x) d x = \frac{1}{7} sin (7 x)

= \frac{1}{2} (sin (x) + \frac{1}{7} sin (7 x))

Add a constant to the solution

= \frac{1}{2} (sin (x) + \frac{1}{7} sin (7 x)) + C

What is the integral of cos(4x)cos(-3x) ?
The integral of cos(4x)cos(-3x) is 1/2 (sin(x)+1/7 sin(7x))+C