What is the integral of 4(tan(x))^{-1} ?

The integral of 4(tan(x))^{-1} is 4ln|sin(x)|+C

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integral of 4(tan(x))^{-1}

\int 4 (tan (x))^{- 1} d x

4 ln ∣ sin (x) ∣ + C

Solution steps

\int 4 (tan (x))^{- 1} d x

Take the constant out:

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

= 4 \cdot \int (tan (x))^{- 1} d x

Rewrite using trig identities

= 4 \cdot \int cot (x) d x

Rewrite using trig identities

= 4 \cdot \int \frac{cos ( x )}{sin ( x )} d x

Apply u-substitution

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

= 4 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 ln ∣ u ∣

Substitute back

u = sin (x)

= 4 ln ∣ sin (x) ∣

Add a constant to the solution

= 4 ln ∣ sin (x) ∣ + C

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What is the integral of 4(tan(x))^{-1} ?
The integral of 4(tan(x))^{-1} is 4ln|sin(x)|+C