What is the integral of tan(kx) ?

The integral of tan(kx) is -1/k ln|cos(kx)|+C

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integral of tan(kx)

\int tan (k x) d x

- \frac{1}{k} ln ∣ cos (k x) ∣ + C

Solution steps

\int tan (k x) d x

Rewrite using trig identities

= \int \frac{sin ( k x )}{cos ( k x )} d x

Apply u-substitution

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

= \int - \frac{1}{k u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{k} ln ∣ u ∣

Substitute back

u = cos (k x)

= - \frac{1}{k} ln ∣ cos (k x) ∣

Add a constant to the solution

= - \frac{1}{k} ln ∣ cos (k x) ∣ + C

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