What is the integral of 2xcot(1-3x^2) ?

The integral of 2xcot(1-3x^2) is -1/3 ln|sin(1-3x^2)|+C

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integral of 2xcot(1-3x^2)

\int 2 x cot (1 - 3 x^{2}) d x

- \frac{1}{3} ln sin (1 - 3 x^{2}) + C

Solution steps

\int 2 x cot (1 - 3 x^{2}) d x

Take the constant out:

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

= 2 \cdot \int x cot (1 - 3 x^{2}) d x

Apply u-substitution

: \int - \frac{cot ( u )}{6} d u

= 2 \cdot \int - \frac{cot ( u )}{6} d u

Take the constant out:

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

= 2 (- \frac{1}{6} \cdot \int cot (u) d u)

Rewrite using trig identities

= 2 (- \frac{1}{6} \cdot \int \frac{cos ( u )}{sin ( u )} d u)

Apply u-substitution

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

= 2 (- \frac{1}{6} \cdot \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 (- \frac{1}{6} ln ∣ v ∣)

Substitute back

= 2 (- \frac{1}{6} ln sin (1 - 3 x^{2}))

Simplify

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

= - \frac{1}{3} ln sin (1 - 3 x^{2})

Add a constant to the solution

= - \frac{1}{3} ln sin (1 - 3 x^{2}) + C

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What is the integral of 2xcot(1-3x^2) ?
The integral of 2xcot(1-3x^2) is -1/3 ln|sin(1-3x^2)|+C