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

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

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integral of 2cos^2(3x)

\int 2 cos^{2} (3 x) d x

x + \frac{1}{6} sin (6 x) + C

Solution steps

\int 2 cos^{2} (3 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int cos^{2} (3 x) d x

Rewrite using trig identities

= 2 \cdot \int \frac{1}{2} (1 + cos (6 x)) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{2} \cdot \int 1 + cos (6 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

= 2 \cdot \frac{1}{2} (\int 1 d x + \int cos (6 x) d x)

\int 1 d x = x

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

= 2 \cdot \frac{1}{2} (x + \frac{1}{6} sin (6 x))

Simplify

2 \cdot \frac{1}{2} (x + \frac{1}{6} sin (6 x)) : x + \frac{1}{6} sin (6 x)

= x + \frac{1}{6} sin (6 x)

Add a constant to the solution

= x + \frac{1}{6} sin (6 x) + C

x \to \infty lim (\frac{9}{2})

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