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

The integral of 1/(cot(y)) is -ln|cos(y)|+C

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integral of 1/(cot(y))

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

- ln ∣ cos (y) ∣ + C

Solution steps

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

Rewrite using trig identities

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

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 (y)

= - ln ∣ cos (y) ∣

Add a constant to the solution

= - ln ∣ cos (y) ∣ + C

x \to 0 - lim (x^{2} - 1)

\int \frac{x ^{2} - 3 x + 2}{x - 1} d x

\int \frac{x ^{2} - 3 x - 7}{( 2 x + 3 ) ( x + 1 ) ^{2}} d x

\frac{d}{d x} (\frac{x ^{2}}{x ^{4} + 1})

\frac{d y}{d x} = 2 y - 1

What is the integral of 1/(cot(y)) ?
The integral of 1/(cot(y)) is -ln|cos(y)|+C