What is the integral of tan^2(6x) ?

The integral of tan^2(6x) is -x+1/6 tan(6x)+C

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integral of tan^2(6x)

\int tan^{2} (6 x) d x

- x + \frac{1}{6} tan (6 x) + C

Solution steps

\int tan^{2} (6 x) d x

Rewrite using trig identities

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

= - \int 1 d x + \int sec^{2} (6 x) d x

\int 1 d x = x

\int sec^{2} (6 x) d x = \frac{1}{6} tan (6 x)

= - x + \frac{1}{6} tan (6 x)

Add a constant to the solution

= - x + \frac{1}{6} tan (6 x) + C

\frac{d y}{d x}, x y = e^{2} y

\frac{m P}{P + ( m - P ) e ^{- km t}}

\frac{d ^{2}}{d x ^{2}} (\frac{3}{4 t ^{5}})

\int \frac{4}{x ^{4} - 1} d x

\frac{d}{d x} (6 x^{4} + 3)

What is the integral of tan^2(6x) ?
The integral of tan^2(6x) is -x+1/6 tan(6x)+C