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

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

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integral of cos^2(3t)

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

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

Solution steps

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

Rewrite using trig identities

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

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 t) d t

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 t + \int cos (6 t) d t)

\int 1 d t = t

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (\frac{x + y}{x ^{2} + y ^{2}})

\int (x^{2} + x - 2) d x

\frac{d}{d x} (arctan (\frac{1}{2} x))

f (x) = 7 - 2 x^{2}

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