What is the integral of tan(x+c) ?

The integral of tan(x+c) is -ln|cos(x+c)|+C

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integral of tan(x+c)

\int tan (x + c) d x

- ln ∣ cos (x + c) ∣ + C

Solution steps

\int tan (x + c) d x

Apply u-substitution

: \int tan (u) d u

= \int tan (u) d u

Rewrite using trig identities

= \int \frac{sin ( u )}{cos ( u )} d u

Apply u-substitution

: \int - \frac{1}{v} d v

= \int - \frac{1}{v} d v

Take the constant out:

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

= - \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - ln ∣ v ∣

Substitute back

= - ln ∣ cos (x + c) ∣

Add a constant to the solution

= - ln ∣ cos (x + c) ∣ + C

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