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

The integral of tan^6(3x) is 1/3 ((tan^5(3x))/5-(tan^3(3x))/3+tan(3x)-arctan(tan(3x)))+C

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integral of tan^6(3x)

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

\frac{1}{3} (\frac{tan ^{5} ( 3 x )}{5} - \frac{tan ^{3} ( 3 x )}{3} + tan (3 x) - arctan (tan (3 x))) + C

Solution steps

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

Apply u-substitution

: \int tan^{6} (u) \frac{1}{3} d u

= \int tan^{6} (u) \frac{1}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int tan^{6} (u) d u

Apply u-substitution

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

= \frac{1}{3} \cdot \int \frac{v ^{6}}{1 + v ^{2}} d v

Long division

\frac{v ^{6}}{1 + v ^{2}} : v^{4} - v^{2} + 1 - \frac{1}{v ^{2} + 1}

= \frac{1}{3} \cdot \int v^{4} - v^{2} + 1 - \frac{1}{v ^{2} + 1} d v

Apply the Sum Rule:

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

= \frac{1}{3} (\int v^{4} d v - \int v^{2} d v + \int 1 d v - \int \frac{1}{v ^{2} + 1} d v)

\int v^{4} d v = \frac{v ^{5}}{5}

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

\int 1 d v = v

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{3} (\frac{v ^{5}}{5} - \frac{v ^{3}}{3} + v - arctan (v))

Substitute back

= \frac{1}{3} (\frac{tan ^{5} ( 3 x )}{5} - \frac{tan ^{3} ( 3 x )}{3} + tan (3 x) - arctan (tan (3 x)))

Add a constant to the solution

= \frac{1}{3} (\frac{tan ^{5} ( 3 x )}{5} - \frac{tan ^{3} ( 3 x )}{3} + tan (3 x) - arctan (tan (3 x))) + C

\int 2 3 x - 1 d x

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\frac{d}{d x} (7 sin (x) + 7 cos (x))

\int e^{6 x} sin (2 x) d x

\int ((x + 1)^{2} (3 - x)^{6}) d x

What is the integral of tan^6(3x) ?
The integral of tan^6(3x) is 1/3 ((tan^5(3x))/5-(tan^3(3x))/3+tan(3x)-arctan(tan(3x)))+C