What is the integral of cot(ax-b) ?

The integral of cot(ax-b) is 1/a ln|sin(ax-b)|+C

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integral of cot(ax-b)

\int cot (a x - b) d x

\frac{1}{a} ln ∣ sin (a x - b) ∣ + C

Solution steps

\int cot (a x - b) d x

Apply u-substitution

: \int \frac{cot ( u )}{a} d u

= \int \frac{cot ( u )}{a} d u

Take the constant out:

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

= \frac{1}{a} \cdot \int cot (u) d u

Rewrite using trig identities

= \frac{1}{a} \cdot \int \frac{cos ( u )}{sin ( u )} d u

Apply u-substitution

: \int \frac{1}{v} d v

= \frac{1}{a} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{a} ln ∣ v ∣

Substitute back

= \frac{1}{a} ln ∣ sin (a x - b) ∣

Add a constant to the solution

= \frac{1}{a} ln ∣ sin (a x - b) ∣ + C

\int 4 sin^{2} (x co) s^{2} x d x

x^{2} + 5 x

x \to 0 + lim (x^{\frac{1}{l n ( x )}})

y = x^{4} + 6 e^{x}, (0, 6)

5 \cdot t \cdot e^{- 5 \cdot t}

What is the integral of cot(ax-b) ?
The integral of cot(ax-b) is 1/a ln|sin(ax-b)|+C