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

The integral of cos(x)cos(7x) is 1/2 (1/6 sin(6x)+1/8 sin(8x))+C

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integral of cos(x)cos(7x)

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

\frac{1}{2} (\frac{1}{6} sin (6 x) + \frac{1}{8} sin (8 x)) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{2} (cos (8 x) + cos (- 6 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 (8 x) + cos (- 6 x) d x

Simplify

cos (8 x) + cos (- 6 x) : cos (6 x) + cos (8 x)

= \frac{1}{2} \cdot \int cos (6 x) + cos (8 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 (6 x) d x + \int cos (8 x) d x)

\int cos (6 x) d x = \frac{1}{6} sin (6 x)

\int cos (8 x) d x = \frac{1}{8} sin (8 x)

= \frac{1}{2} (\frac{1}{6} sin (6 x) + \frac{1}{8} sin (8 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{6} sin (6 x) + \frac{1}{8} sin (8 x)) + C

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

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

\frac{d}{d x} (- 8 ln (x))

y = 7 x^{- 7}

What is the integral of cos(x)cos(7x) ?
The integral of cos(x)cos(7x) is 1/2 (1/6 sin(6x)+1/8 sin(8x))+C