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

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

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

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

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

Solution steps

\int cos (4 x) cos (6 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

Simplify

cos (10 x) + cos (- 2 x) : cos (2 x) + cos (10 x)

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

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

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

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

Add a constant to the solution

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

\int \frac{1}{( x ^{2} + R ^{2} ) ^{\frac{3}{2}}} d x

2 y + x y^{^{'}} = 0

\frac{d}{d x} (\frac{2}{x ^{3}} - \frac{3}{x ^{2}})

\frac{d}{d x} (\frac{2}{( x ^{2} + x + 1 ) ^{2}})

\frac{\partial}{\partial y} (2 y + 1)

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