What is the integral of 1/(sqrt(tan^2(x))) ?

The integral of 1/(sqrt(tan^2(x))) is ln|sin(x)|+C

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integral of 1/(sqrt(tan^2(x)))

\int \frac{1}{tan ^{2} ( x )} d x

ln ∣ sin (x) ∣ + C

Solution steps

\int \frac{1}{tan ^{2} ( x )} d x

tan^{2} (x) = (tan (x)),

assuming

tan (x) \geq 0

= \int \frac{1}{tan ( 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

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What is the integral of 1/(sqrt(tan^2(x))) ?
The integral of 1/(sqrt(tan^2(x))) is ln|sin(x)|+C