What is the integral of cos(x)cos(2x)cos(3x) ?

The integral of cos(x)cos(2x)cos(3x) is

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integral of cos(x)cos(2x)cos(3x)

\int cos (x) cos (2 x) cos (3 x) d x

\frac{1}{2} cos (x) (sin (x) + \frac{1}{5} sin (5 x)) + \frac{1}{2} (\frac{1}{2} (x - \frac{1}{2} sin (2 x)) + \frac{1}{10} (\frac{1}{4} sin (4 x) - \frac{1}{6} sin (6 x))) + C

Solution steps

\int cos (x) cos (2 x) cos (3 x) d x

Apply Integration By Parts

: \frac{1}{2} cos (x) (sin (x) + \frac{1}{5} sin (5 x)) - \int - \frac{1}{2} sin (x) (sin (x) + \frac{1}{5} sin (5 x)) d x

= \frac{1}{2} cos (x) (sin (x) + \frac{1}{5} sin (5 x)) - \int - \frac{1}{2} sin (x) (sin (x) + \frac{1}{5} sin (5 x)) d x

\int - \frac{1}{2} sin (x) (sin (x) + \frac{1}{5} sin (5 x)) d x = - \frac{1}{2} (\frac{1}{2} (x - \frac{1}{2} sin (2 x)) + \frac{1}{10} (\frac{1}{4} sin (4 x) - \frac{1}{6} sin (6 x)))

= \frac{1}{2} cos (x) (sin (x) + \frac{1}{5} sin (5 x)) - (- \frac{1}{2} (\frac{1}{2} (x - \frac{1}{2} sin (2 x)) + \frac{1}{10} (\frac{1}{4} sin (4 x) - \frac{1}{6} sin (6 x))))

Simplify

= \frac{1}{2} cos (x) (sin (x) + \frac{1}{5} sin (5 x)) + \frac{1}{2} (\frac{1}{2} (x - \frac{1}{2} sin (2 x)) + \frac{1}{10} (\frac{1}{4} sin (4 x) - \frac{1}{6} sin (6 x)))

Add a constant to the solution

= \frac{1}{2} cos (x) (sin (x) + \frac{1}{5} sin (5 x)) + \frac{1}{2} (\frac{1}{2} (x - \frac{1}{2} sin (2 x)) + \frac{1}{10} (\frac{1}{4} sin (4 x) - \frac{1}{6} sin (6 x))) + C

\int_{0}^{\frac{3}{2} 3} \frac{x ^{3}}{9 - x ^{2}} d x

y = \frac{e ^{x} - e ^{- x}}{2}

\int x^{3} + 3 x - 2 d x

\frac{d}{d t} (\frac{1}{3} t^{3} + \frac{1}{2} t^{2})

\frac{d}{d x} (e^{- 8 x^{2}})

What is the integral of cos(x)cos(2x)cos(3x) ?
The integral of cos(x)cos(2x)cos(3x) is