What is the integral of cot(6x) ?

The integral of cot(6x) is 1/6 ln|sin(6x)|+C

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integral of cot(6x)

\int cot (6 x) d x

\frac{1}{6} ln ∣ sin (6 x) ∣ + C

Solution steps

\int cot (6 x) d x

Rewrite using trig identities

= \int \frac{cos ( 6 x )}{sin ( 6 x )} d x

Apply u-substitution

: \int \frac{1}{6 u} d u

= \int \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = sin (6 x)

= \frac{1}{6} ln ∣ sin (6 x) ∣

Add a constant to the solution

= \frac{1}{6} ln ∣ sin (6 x) ∣ + C

\frac{\partial}{\partial z} (x^{3})

x^{2} sin^{2} (5 x)

y = 5 e^{x}

\int \frac{x}{x - 1} d x

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