What is the integral of tan^5(3x) ?

The integral of tan^5(3x) is 1/3 ln|sec(3x)|-(sec^2(3x))/3+(sec^4(3x))/(12)+C

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integral of tan^5(3x)

\int tan^{5} (3 x) d x

\frac{1}{3} ln ∣ sec (3 x) ∣ - \frac{sec ^{2} ( 3 x )}{3} + \frac{sec ^{4} ( 3 x )}{12} + C

Solution steps

\int tan^{5} (3 x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{( - 1 + u ^{2} ) ^{2}}{3 u} d u

= \int \frac{( - 1 + u ^{2} ) ^{2}}{3 u} d u

Expand

\frac{( - 1 + u ^{2} ) ^{2}}{3 u} : \frac{1}{3 u} - \frac{2 u}{3} + \frac{u ^{3}}{3}

= \int \frac{1}{3 u} - \frac{2 u}{3} + \frac{u ^{3}}{3} 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}{3 u} d u - \int \frac{2 u}{3} d u + \int \frac{u ^{3}}{3} d u

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

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

\int \frac{u ^{3}}{3} d u = \frac{u ^{4}}{12}

= \frac{1}{3} ln ∣ u ∣ - \frac{u ^{2}}{3} + \frac{u ^{4}}{12}

Substitute back

u = sec (3 x)

= \frac{1}{3} ln ∣ sec (3 x) ∣ - \frac{sec ^{2} ( 3 x )}{3} + \frac{sec ^{4} ( 3 x )}{12}

Add a constant to the solution

= \frac{1}{3} ln ∣ sec (3 x) ∣ - \frac{sec ^{2} ( 3 x )}{3} + \frac{sec ^{4} ( 3 x )}{12} + C

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What is the integral of tan^5(3x) ?
The integral of tan^5(3x) is 1/3 ln|sec(3x)|-(sec^2(3x))/3+(sec^4(3x))/(12)+C