What is the integral of tan^4(3x) ?

The integral of tan^4(3x) is 1/9 tan^3(3x)+x-1/3 tan(3x)+C

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integral of tan^4(3x)

\int tan^{4} (3 x) d x

\frac{1}{9} tan^{3} (3 x) + x - \frac{1}{3} tan (3 x) + C

Solution steps

\int tan^{4} (3 x) d x

Apply u-substitution

: \int tan^{4} (u) \frac{1}{3} d u

= \int tan^{4} (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int tan^{4} (u) d u

Apply Integral Reduction

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

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

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

Substitute back

u = 3 x

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

Simplify

\frac{1}{3} (\frac{tan ^{3} ( 3 x )}{3} - (- 3 x + tan (3 x))) : \frac{1}{9} tan^{3} (3 x) + x - \frac{1}{3} tan (3 x)

= \frac{1}{9} tan^{3} (3 x) + x - \frac{1}{3} tan (3 x)

Add a constant to the solution

= \frac{1}{9} tan^{3} (3 x) + x - \frac{1}{3} tan (3 x) + C

\frac{\partial}{\partial x} (ln (x yz))

f (x) = \frac{3}{4 x + 7}

\frac{d}{d x} (\frac{2 x - 5}{x + 1})

x \to 0 lim (2 x cot (4 x))

y = \frac{6 x}{x + 2}

What is the integral of tan^4(3x) ?
The integral of tan^4(3x) is 1/9 tan^3(3x)+x-1/3 tan(3x)+C