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

The integral of cos(6x)cos(x)sin(4x)sin(x) is

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

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

\frac{1}{2} (\frac{1}{2} cos (6 x) (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) + 3 (\frac{1}{4} (- \frac{1}{8} sin (8 x) + \frac{1}{4} sin (4 x)) - \frac{1}{12} (x - \frac{1}{12} sin (12 x)))) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{2} sin (2 x) cos (6 x) sin (4 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 sin (2 x) cos (6 x) sin (4 x) d x

Apply Integration By Parts

: \frac{1}{2} cos (6 x) (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) - \int - 3 sin (6 x) (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) d x

= \frac{1}{2} (\frac{1}{2} cos (6 x) (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) - \int - 3 sin (6 x) (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) d x)

\int - 3 sin (6 x) (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) d x = - 3 (\frac{1}{4} (- \frac{1}{8} sin (8 x) + \frac{1}{4} sin (4 x)) - \frac{1}{12} (x - \frac{1}{12} sin (12 x)))

= \frac{1}{2} (\frac{1}{2} cos (6 x) (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) - (- 3 (\frac{1}{4} (- \frac{1}{8} sin (8 x) + \frac{1}{4} sin (4 x)) - \frac{1}{12} (x - \frac{1}{12} sin (12 x)))))

Simplify

= \frac{1}{2} (\frac{1}{2} cos (6 x) (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) + 3 (\frac{1}{4} (- \frac{1}{8} sin (8 x) + \frac{1}{4} sin (4 x)) - \frac{1}{12} (x - \frac{1}{12} sin (12 x))))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{2} cos (6 x) (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) + 3 (\frac{1}{4} (- \frac{1}{8} sin (8 x) + \frac{1}{4} sin (4 x)) - \frac{1}{12} (x - \frac{1}{12} sin (12 x)))) + C

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