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

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

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integral of cos^2(3x)

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

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

Solution steps

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

Rewrite using trig identities

= \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

= \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

= \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)

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

Add a constant to the solution

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

f (x) = arcsin (2 x + 1)

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

\frac{d}{d x} (\frac{x - 1}{x - 1})

\int 16 sin^{3} (x co) s^{2} x d x

\int sin (x t) d t

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