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

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

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integral of tan^4(6x)

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

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

Solution steps

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

Apply u-substitution

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

= \int tan^{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 tan^{4} (u) d u

Apply Integral Reduction

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

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

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

Substitute back

u = 6 x

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

Simplify

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

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

Add a constant to the solution

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

5 x + 3

\int \frac{10}{x - 5} d x

f (x) = 2 x^{2} - 7 x + 5, at x = 3

\int_{0}^{t} cos (x^{2}) d x

\frac{\partial}{\partial x} (cos (4 θ))

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