What is the integral of tan(3θ) ?

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

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integral of tan(3θ)

\int tan (3 θ) d θ

- \frac{1}{3} ln ∣ cos (3 θ) ∣ + C

Solution steps

\int tan (3 θ) d θ

Rewrite using trig identities

= \int \frac{sin ( 3 θ )}{cos ( 3 θ )} d θ

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{3} ln ∣ u ∣

Substitute back

u = cos (3 θ)

= - \frac{1}{3} ln ∣ cos (3 θ) ∣

Add a constant to the solution

= - \frac{1}{3} ln ∣ cos (3 θ) ∣ + C

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