What is the integral of 1/(tan^4(6x)) ?

The integral of 1/(tan^4(6x)) is -1/18 cot^3(6x)+x+1/6 cot(6x)+C

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integral of 1/(tan^4(6x))

\int \frac{1}{tan ^{4} ( 6 x )} d x

- \frac{1}{18} cot^{3} (6 x) + x + \frac{1}{6} cot (6 x) + C

Solution steps

\int \frac{1}{tan ^{4} ( 6 x )} d x

Use the basic trigonometric identity:

\frac{1}{tan ( x )} = cot (x)

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

Apply u-substitution

= \int cot^{4} (u) \frac{1}{6} 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 cot^{4} (u) d u

Apply Integral Reduction

= \frac{1}{6} (- \frac{cot ^{3} ( u )}{3} - \int cot^{2} (u) d u)

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

= \frac{1}{6} (- \frac{cot ^{3} ( u )}{3} - (- u - cot (u)))

Substitute back

u = 6 x

= \frac{1}{6} (- \frac{cot ^{3} ( 6 x )}{3} - (- 6 x - cot (6 x)))

Simplify

\frac{1}{6} (- \frac{cot ^{3} ( 6 x )}{3} - (- 6 x - cot (6 x))) : - \frac{1}{18} cot^{3} (6 x) + x + \frac{1}{6} cot (6 x)

= - \frac{1}{18} cot^{3} (6 x) + x + \frac{1}{6} cot (6 x)

Add a constant to the solution

= - \frac{1}{18} cot^{3} (6 x) + x + \frac{1}{6} cot (6 x) + C

What is the integral of 1/(tan^4(6x)) ?
The integral of 1/(tan^4(6x)) is -1/18 cot^3(6x)+x+1/6 cot(6x)+C