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

The integral of cot^2(6x^2)x is 1/2 (-x^2-1/6 cot(6x^2))+C

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

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

\frac{1}{2} (- x^{2} - \frac{1}{6} cot (6 x^{2})) + C

Solution steps

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

Apply u-substitution

: \int \frac{cot ^{2} ( 6 u )}{2} d u

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

Take the constant out:

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

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

Rewrite using trig identities

= \frac{1}{2} \cdot \int - 1 + csc^{2} (6 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- \int 1 d u + \int csc^{2} (6 u) d u)

\int 1 d u = u

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

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

Substitute back

u = x^{2}

= \frac{1}{2} (- x^{2} - \frac{1}{6} cot (6 x^{2}))

Add a constant to the solution

= \frac{1}{2} (- x^{2} - \frac{1}{6} cot (6 x^{2})) + C

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What is the integral of cot^2(6x^2)x ?
The integral of cot^2(6x^2)x is 1/2 (-x^2-1/6 cot(6x^2))+C