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

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

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integral of 1/(tan(y))

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

ln ∣ sin (y) ∣ + C

Solution steps

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

Rewrite using trig identities

= \int \frac{cos ( y )}{sin ( y )} d y

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 (y)

= ln ∣ sin (y) ∣

Add a constant to the solution

= ln ∣ sin (y) ∣ + C

y^{^{''}} - 8 y^{^{'}} + 16 y = t^{- 7} e^{4 t}

x \to 1 + lim (x^{2})

x \to 5 lim (\frac{x}{x - 5})

\int \frac{1}{y ^{\frac{1}{3}}} d y

(x - 1) y^{^{'}} + (2 x + 1) y = 0