What is the integral of cos(x)cos(3x)cos(5x) ?

The integral of cos(x)cos(3x)cos(5x) is

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integral of cos(x)cos(3x)cos(5x)

\int cos (x) cos (3 x) cos (5 x) d x

\frac{1}{2} cos (x) (\frac{1}{2} sin (2 x) + \frac{1}{8} sin (8 x)) + \frac{1}{2} (\frac{1}{3} sin^{3} (x) + \frac{1}{16} (\frac{1}{7} sin (7 x) - \frac{1}{9} sin (9 x))) + C

Solution steps

\int cos (x) cos (3 x) cos (5 x) d x

Apply Integration By Parts

: \frac{1}{2} cos (x) (\frac{1}{2} sin (2 x) + \frac{1}{8} sin (8 x)) - \int - \frac{1}{2} sin (x) (\frac{1}{2} sin (2 x) + \frac{1}{8} sin (8 x)) d x

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

\int - \frac{1}{2} sin (x) (\frac{1}{2} sin (2 x) + \frac{1}{8} sin (8 x)) d x = - \frac{1}{2} (\frac{1}{3} sin^{3} (x) + \frac{1}{16} (\frac{1}{7} sin (7 x) - \frac{1}{9} sin (9 x)))

= \frac{1}{2} cos (x) (\frac{1}{2} sin (2 x) + \frac{1}{8} sin (8 x)) - (- \frac{1}{2} (\frac{1}{3} sin^{3} (x) + \frac{1}{16} (\frac{1}{7} sin (7 x) - \frac{1}{9} sin (9 x))))

Simplify

= \frac{1}{2} cos (x) (\frac{1}{2} sin (2 x) + \frac{1}{8} sin (8 x)) + \frac{1}{2} (\frac{1}{3} sin^{3} (x) + \frac{1}{16} (\frac{1}{7} sin (7 x) - \frac{1}{9} sin (9 x)))

Add a constant to the solution

= \frac{1}{2} cos (x) (\frac{1}{2} sin (2 x) + \frac{1}{8} sin (8 x)) + \frac{1}{2} (\frac{1}{3} sin^{3} (x) + \frac{1}{16} (\frac{1}{7} sin (7 x) - \frac{1}{9} sin (9 x))) + C

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What is the integral of cos(x)cos(3x)cos(5x) ?
The integral of cos(x)cos(3x)cos(5x) is