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

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

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

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

y^{^{'}} = \frac{( 6 x ^{4} e ^{\frac{y}{x}} + x ^{2} y ^{2} )}{x ^{3} y}

n = 0 \sum \infty \frac{1}{( n ^{2} - 1 )}

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

x \to 1 - lim (- \frac{2}{1 - x})

x \to 0 lim (ln (sin (x)))

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