What is the integral of tan(-1x) ?

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

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integral of tan(-1x)

\int tan (- 1 x) d x

ln ∣ cos (x) ∣ + C

Solution steps

\int tan (- 1 \cdot x) d x

Simplify

tan (- 1 \cdot x) : - tan (x)

= \int - tan (x) d x

Take the constant out:

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

= - \int tan (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- ln ∣ u ∣)

Substitute back

u = cos (x)

= - (- ln ∣ cos (x) ∣)

Simplify

= ln ∣ cos (x) ∣

Add a constant to the solution

= ln ∣ cos (x) ∣ + C

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