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

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

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integral of 1/(tan(x))

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

ln ∣ sin (x) ∣ + C

Solution steps

\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

(x + 1)^{2} y^{^{'}} + 5 (x + 1) y = 16

i = 1 \sum \infty \frac{1}{i ^{2}}

\int e^{- 2 a x} d x

\int (\frac{x ^{2} + 2 x - 3}{x ^{4}}) d x

x = e^{y}, y = 0, x = 1, x = e^{2}