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

The integral of 2cos(x)-sin(3x)cos(3x) is 2sin(x)+1/6 cos^2(3x)+C

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

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

2 sin (x) + \frac{1}{6} cos^{2} (3 x) + C

Solution steps

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

= \int 2 cos (x) d x - \int sin (3 x) cos (3 x) d x

\int 2 cos (x) d x = 2 sin (x)

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

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

Simplify

= 2 sin (x) + \frac{1}{6} cos^{2} (3 x)

Add a constant to the solution

= 2 sin (x) + \frac{1}{6} cos^{2} (3 x) + C

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What is the integral of 2cos(x)-sin(3x)cos(3x) ?
The integral of 2cos(x)-sin(3x)cos(3x) is 2sin(x)+1/6 cos^2(3x)+C