What is the integral of-tan(x) ?

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

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integral of-tan(x)

\int - tan (x) d x

ln ∣ cos (x) ∣ + C

Solution steps

\int - tan (x) d x

Take the constant out:

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

= - \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) ∣)

Simplify

= ln ∣ cos (x) ∣

Add a constant to the solution

= ln ∣ cos (x) ∣ + C

\frac{d y}{d x} = x^{2} y

x y + (x^{2} + y^{2}) (\frac{d y}{d x}) = 0

s \to - 3 lim (3 \frac{- 4 + 5 s + 3 s ^{2}}{- s ^{3} + s + 40})

f (x) = sin (x)

(\frac{1}{x + 2})^{^{'}}