What is the integral of cos(4x)cos(7x) ?

The integral of cos(4x)cos(7x) is 1/2 (1/3 sin(3x)+1/11 sin(11x))+C

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integral of cos(4x)cos(7x)

\int cos (4 x) cos (7 x) d x

\frac{1}{2} (\frac{1}{3} sin (3 x) + \frac{1}{11} sin (11 x)) + C

Solution steps

\int cos (4 x) cos (7 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (cos (11 x) + cos (- 3 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 (11 x) + cos (- 3 x) d x

Simplify

cos (11 x) + cos (- 3 x) : cos (3 x) + cos (11 x)

= \frac{1}{2} \cdot \int cos (3 x) + cos (11 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 (3 x) d x + \int cos (11 x) d x)

\int cos (3 x) d x = \frac{1}{3} sin (3 x)

\int cos (11 x) d x = \frac{1}{11} sin (11 x)

= \frac{1}{2} (\frac{1}{3} sin (3 x) + \frac{1}{11} sin (11 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{3} sin (3 x) + \frac{1}{11} sin (11 x)) + C

f (x) = - \frac{6}{x ^{4}}

n \to \infty lim (n 10 n)

3 y^{^{'}} + y = 0

\frac{\partial}{\partial z} (x^{2} + 2 y^{2} + 3 z^{2})

\frac{\partial}{\partial x} (e^{- 5 x^{2} - y^{2}})

What is the integral of cos(4x)cos(7x) ?
The integral of cos(4x)cos(7x) is 1/2 (1/3 sin(3x)+1/11 sin(11x))+C