What is the integral of cot(16x) ?

The integral of cot(16x) is 1/16 ln|sin(16x)|+C

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integral of cot(16x)

\int cot (16 x) d x

\frac{1}{16} ln ∣ sin (16 x) ∣ + C

Solution steps

\int cot (16 x) d x

Rewrite using trig identities

= \int \frac{cos ( 16 x )}{sin ( 16 x )} d x

Apply u-substitution

: \int \frac{1}{16 u} d u

= \int \frac{1}{16 u} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{16} ln ∣ u ∣

Substitute back

u = sin (16 x)

= \frac{1}{16} ln ∣ sin (16 x) ∣

Add a constant to the solution

= \frac{1}{16} ln ∣ sin (16 x) ∣ + C

x \to 0 - lim (\frac{5}{x ^{3}})

x^{18}

\frac{d}{d x} (tan (2 x) cot (2 x))

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What is the integral of cot(16x) ?
The integral of cot(16x) is 1/16 ln|sin(16x)|+C