What is the integral of tan^3(6x) ?

The integral of tan^3(6x) is -1/6 ln|sec(6x)|+(sec^2(6x))/(12)+C

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integral of tan^3(6x)

\int tan^{3} (6 x) d x

- \frac{1}{6} ln ∣ sec (6 x) ∣ + \frac{sec ^{2} ( 6 x )}{12} + C

Solution steps

\int tan^{3} (6 x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{- 1 + u ^{2}}{6 u} d u

= \int \frac{- 1 + u ^{2}}{6 u} d u

Expand

\frac{- 1 + u ^{2}}{6 u} : - \frac{1}{6 u} + \frac{u}{6}

= \int - \frac{1}{6 u} + \frac{u}{6} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{6 u} d u + \int \frac{u}{6} d u

\int \frac{1}{6 u} d u = \frac{1}{6} ln ∣ u ∣

\int \frac{u}{6} d u = \frac{u ^{2}}{12}

= - \frac{1}{6} ln ∣ u ∣ + \frac{u ^{2}}{12}

Substitute back

u = sec (6 x)

= - \frac{1}{6} ln ∣ sec (6 x) ∣ + \frac{sec ^{2} ( 6 x )}{12}

Add a constant to the solution

= - \frac{1}{6} ln ∣ sec (6 x) ∣ + \frac{sec ^{2} ( 6 x )}{12} + C

\int \frac{x}{x ^{2} - 6 x + 5} d x

\int (7 + x - 3 x^{2}) e^{- x} d x

sin (x)

2 x^{3} + 3 x^{2} - 12 x + 8

x \to \infty lim (e^{n x})

What is the integral of tan^3(6x) ?
The integral of tan^3(6x) is -1/6 ln|sec(6x)|+(sec^2(6x))/(12)+C