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

The integral of tan^3(6x)sec(6x) is -1/6 sec(6x)+(sec^3(6x))/(18)+C

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integral of tan^3(6x)sec(6x)

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

- \frac{1}{6} sec (6 x) + \frac{sec ^{3} ( 6 x )}{18} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

\int \frac{u ^{2}}{6} d u = \frac{u ^{3}}{18}

= - \frac{1}{6} u + \frac{u ^{3}}{18}

Substitute back

u = sec (6 x)

= - \frac{1}{6} sec (6 x) + \frac{sec ^{3} ( 6 x )}{18}

Add a constant to the solution

= - \frac{1}{6} sec (6 x) + \frac{sec ^{3} ( 6 x )}{18} + C

\int y^{2} - x^{2} d x

x \to - 2 lim (25 - 5 x)

f (x) = x^{\frac{4}{3}} - 4 x^{\frac{1}{3}} + 1

\int_{0}^{π} sin (θ) d θ

\frac{d}{d x} (x^{2} 5 - x)

What is the integral of tan^3(6x)sec(6x) ?
The integral of tan^3(6x)sec(6x) is -1/6 sec(6x)+(sec^3(6x))/(18)+C