What is the integral of cos(6x)cos(4x) ?

The integral of cos(6x)cos(4x) is 1/2 (1/10 sin(10x)+1/2 sin(2x))+C

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integral of cos(6x)cos(4x)

\int cos (6 x) cos (4 x) d x

\frac{1}{2} (\frac{1}{10} sin (10 x) + \frac{1}{2} sin (2 x)) + C

Solution steps

\int cos (6 x) cos (4 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (cos (10 x) + cos (2 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 (10 x) + cos (2 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 (10 x) d x + \int cos (2 x) d x)

\int cos (10 x) d x = \frac{1}{10} sin (10 x)

\int cos (2 x) d x = \frac{1}{2} sin (2 x)

= \frac{1}{2} (\frac{1}{10} sin (10 x) + \frac{1}{2} sin (2 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{10} sin (10 x) + \frac{1}{2} sin (2 x)) + C

x (x - 9)^{2}

\int_{2}^{4} 5 x^{2} d x

f (x) = 4 - x^{2}, at x = - 4

\int \frac{10}{x} + 10 x d x

x \to - 2 + lim (\frac{∣ x + 2 ∣}{x + 2})

What is the integral of cos(6x)cos(4x) ?
The integral of cos(6x)cos(4x) is 1/2 (1/10 sin(10x)+1/2 sin(2x))+C