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

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

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integral of tan(c-x)

\int tan (c - x) d x

ln ∣ cos (c - x) ∣ + C

Solution steps

\int tan (c - x) d x

Apply u-substitution

: \int - tan (u) d u

= \int - tan (u) d u

Take the constant out:

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

= - \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 (c - x) ∣)

Simplify

= ln ∣ cos (c - x) ∣

Add a constant to the solution

= ln ∣ cos (c - x) ∣ + C

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