What is the integral of-tan(t) ?

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

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integral of-tan(t)

\int - tan (t) d t

ln ∣ cos (t) ∣ + C

Solution steps

\int - tan (t) d t

Take the constant out:

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

= - \int tan (t) d t

Rewrite using trig identities

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

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 (t)

= - (- ln ∣ cos (t) ∣)

Simplify

= ln ∣ cos (t) ∣

Add a constant to the solution

= ln ∣ cos (t) ∣ + C

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