What is the integral of (tan(3x))/3 ?

The integral of (tan(3x))/3 is -1/9 ln|cos(3x)|+C

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integral of (tan(3x))/3

\int \frac{tan ( 3 x )}{3} d x

- \frac{1}{9} ln ∣ cos (3 x) ∣ + C

Solution steps

\int \frac{tan ( 3 x )}{3} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int tan (3 x) d x

Rewrite using trig identities

= \frac{1}{3} \cdot \int \frac{sin ( 3 x )}{cos ( 3 x )} d x

Apply u-substitution

: \int - \frac{1}{3 u} d u

= \frac{1}{3} \cdot \int - \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} (- \frac{1}{3} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} (- \frac{1}{3} ln ∣ u ∣)

Substitute back

u = cos (3 x)

= \frac{1}{3} (- \frac{1}{3} ln ∣ cos (3 x) ∣)

Simplify

\frac{1}{3} (- \frac{1}{3} ln ∣ cos (3 x) ∣) : - \frac{1}{9} ln ∣ cos (3 x) ∣

= - \frac{1}{9} ln ∣ cos (3 x) ∣

Add a constant to the solution

= - \frac{1}{9} ln ∣ cos (3 x) ∣ + C

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What is the integral of (tan(3x))/3 ?
The integral of (tan(3x))/3 is -1/9 ln|cos(3x)|+C