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

The integral of tan^2(3t) is -t+1/3 tan(3t)+C

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integral of tan^2(3t)

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

- t + \frac{1}{3} tan (3 t) + C

Solution steps

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

Rewrite using trig identities

= \int - 1 + sec^{2} (3 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

= - \int 1 d t + \int sec^{2} (3 t) d t

\int 1 d t = t

\int sec^{2} (3 t) d t = \frac{1}{3} tan (3 t)

= - t + \frac{1}{3} tan (3 t)

Add a constant to the solution

= - t + \frac{1}{3} tan (3 t) + C

\frac{\partial}{\partial x} (x \cdot a + b)

y = \frac{x - 1}{x - 2} + 1, (3, 3)

ln (x^{2} + y) - x y^{2} = 10

\frac{cos ( x )}{1 + a \cdot sin ( x )}

\int 14 e^{- 14 x} d x

What is the integral of tan^2(3t) ?
The integral of tan^2(3t) is -t+1/3 tan(3t)+C