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

The integral of (tan(x))^{-1} is ln|sin(x)|+C

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integral of (tan(x))^{-1}

\int (tan (x))^{- 1} d x

ln ∣ sin (x) ∣ + C

Solution steps

\int (tan (x))^{- 1} d x

Rewrite using trig identities

= \int cot (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = sin (x)

= ln ∣ sin (x) ∣

Add a constant to the solution

= ln ∣ sin (x) ∣ + C

\frac{\partial}{\partial x} (lo g_{10} (x))

x^{^{'}} = x - 4

x \to \infty lim (- cos (x))

\frac{d}{d x} (cos (x) + \frac{1}{x})

\frac{\partial}{\partial x} (2 cos (π x))

What is the integral of (tan(x))^{-1} ?
The integral of (tan(x))^{-1} is ln|sin(x)|+C