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

The integral of tan^7(3x) is -1/3 ln|sec(3x)|+(sec^2(3x))/2-(sec^4(3x))/4+(sec^6(3x))/(18)+C

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integral of tan^7(3x)

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

- \frac{1}{3} ln ∣ sec (3 x) ∣ + \frac{sec ^{2} ( 3 x )}{2} - \frac{sec ^{4} ( 3 x )}{4} + \frac{sec ^{6} ( 3 x )}{18} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

\frac{( - 1 + u ^{2} ) ^{3}}{3 u} : - \frac{1}{3 u} + u - u^{3} + \frac{u ^{5}}{3}

= \int - \frac{1}{3 u} + u - u^{3} + \frac{u ^{5}}{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 u d u - \int u^{3} d u + \int \frac{u ^{5}}{3} d u

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

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

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

\int \frac{u ^{5}}{3} d u = \frac{u ^{6}}{18}

= - \frac{1}{3} ln ∣ u ∣ + \frac{u ^{2}}{2} - \frac{u ^{4}}{4} + \frac{u ^{6}}{18}

Substitute back

u = sec (3 x)

= - \frac{1}{3} ln ∣ sec (3 x) ∣ + \frac{sec ^{2} ( 3 x )}{2} - \frac{sec ^{4} ( 3 x )}{4} + \frac{sec ^{6} ( 3 x )}{18}

Add a constant to the solution

= - \frac{1}{3} ln ∣ sec (3 x) ∣ + \frac{sec ^{2} ( 3 x )}{2} - \frac{sec ^{4} ( 3 x )}{4} + \frac{sec ^{6} ( 3 x )}{18} + C

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What is the integral of tan^7(3x) ?
The integral of tan^7(3x) is -1/3 ln|sec(3x)|+(sec^2(3x))/2-(sec^4(3x))/4+(sec^6(3x))/(18)+C