What is the integral of tan^3(7x) ?

The integral of tan^3(7x) is -1/7 ln|sec(7x)|+(sec^2(7x))/(14)+C

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integral of tan^3(7x)

\int tan^{3} (7 x) d x

- \frac{1}{7} ln ∣ sec (7 x) ∣ + \frac{sec ^{2} ( 7 x )}{14} + C

Solution steps

\int tan^{3} (7 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (7 x)) tan (7 x) d x

Apply u-substitution

: \int \frac{- 1 + u ^{2}}{7 u} d u

= \int \frac{- 1 + u ^{2}}{7 u} d u

Expand

\frac{- 1 + u ^{2}}{7 u} : - \frac{1}{7 u} + \frac{u}{7}

= \int - \frac{1}{7 u} + \frac{u}{7} 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}{7 u} d u + \int \frac{u}{7} d u

\int \frac{1}{7 u} d u = \frac{1}{7} ln ∣ u ∣

\int \frac{u}{7} d u = \frac{u ^{2}}{14}

= - \frac{1}{7} ln ∣ u ∣ + \frac{u ^{2}}{14}

Substitute back

u = sec (7 x)

= - \frac{1}{7} ln ∣ sec (7 x) ∣ + \frac{sec ^{2} ( 7 x )}{14}

Add a constant to the solution

= - \frac{1}{7} ln ∣ sec (7 x) ∣ + \frac{sec ^{2} ( 7 x )}{14} + C

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What is the integral of tan^3(7x) ?
The integral of tan^3(7x) is -1/7 ln|sec(7x)|+(sec^2(7x))/(14)+C