What is the integral of-4cot(6x^2)2x ?

The integral of-4cot(6x^2)2x is -2/3 ln|sin(6x^2)|+C

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integral of-4cot(6x^2)2x

\int - 4 cot (6 x^{2}) 2 x d x

- \frac{2}{3} ln sin (6 x^{2}) + C

Solution steps

\int - 4 cot (6 x^{2}) \cdot 2 x d x

Take the constant out:

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

= - 4 \cdot 2 \cdot \int cot (6 x^{2}) x d x

Simplify

= - 8 \cdot \int cot (6 x^{2}) x d x

Apply u-substitution

= - 8 \cdot \int \frac{cot ( u )}{12} d u

Take the constant out:

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

= - 8 \cdot \frac{1}{12} \cdot \int cot (u) d u

Rewrite using trig identities

= - 8 \cdot \frac{1}{12} \cdot \int \frac{cos ( u )}{sin ( u )} d u

Apply u-substitution

= - 8 \cdot \frac{1}{12} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - 8 \cdot \frac{1}{12} ln ∣ v ∣

Substitute back

= - 8 \cdot \frac{1}{12} ln sin (6 x^{2})

Simplify

- 8 \cdot \frac{1}{12} ln sin (6 x^{2}) : - \frac{2}{3} ln sin (6 x^{2})

= - \frac{2}{3} ln sin (6 x^{2})

Add a constant to the solution

= - \frac{2}{3} ln sin (6 x^{2}) + C

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What is the integral of-4cot(6x^2)2x ?
The integral of-4cot(6x^2)2x is -2/3 ln|sin(6x^2)|+C