What is the integral of tan^2(3x-1) ?

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

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integral of tan^2(3x-1)

\int tan^{2} (3 x - 1) d x

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

Solution steps

\int tan^{2} (3 x - 1) d x

Rewrite using trig identities

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

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

\int 1 d x = x

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

= - x - \frac{1}{3} tan (1 - 3 x)

Add a constant to the solution

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

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\int 216 x^{2} cos (6 x) d x

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