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

The integral of sqrt(tan^2(x)) is -ln|cos(x)|+C

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integral of sqrt(tan^2(x))

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

- ln ∣ cos (x) ∣ + C

Solution steps

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

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

assuming

tan (x) \geq 0

= \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) ∣

Add a constant to the solution

= - ln ∣ cos (x) ∣ + C

\int \frac{ln ( q )}{q ^{2}} d q

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\int [\frac{x ^{2} - 9}{x}] d x

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

\int (2 x + \frac{1}{2 x}) d x

What is the integral of sqrt(tan^2(x)) ?
The integral of sqrt(tan^2(x)) is -ln|cos(x)|+C