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

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

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integral of (cos(3x))^2

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

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

Solution steps

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

Simplify

= \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

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 t} + 25 e^{t} cos (3 t)

0.125 y^{^{''}} + 4 y^{^{'}} + 28 y = 0, y (0) = 0

\int 2 x (x^{2} + 9)^{6} d x

\frac{d}{d x} (x^{4} - 6 x^{2})

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