What is the integral of tan^4(9x) ?

The integral of tan^4(9x) is 1/27 tan^3(9x)+x-1/9 tan(9x)+C

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integral of tan^4(9x)

\int tan^{4} (9 x) d x

\frac{1}{27} tan^{3} (9 x) + x - \frac{1}{9} tan (9 x) + C

Solution steps

\int tan^{4} (9 x) d x

Apply u-substitution

: \int tan^{4} (u) \frac{1}{9} d u

= \int tan^{4} (u) \frac{1}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int tan^{4} (u) d u

Apply Integral Reduction

= \frac{1}{9} (\frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u)

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

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

Substitute back

u = 9 x

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

Simplify

\frac{1}{9} (\frac{tan ^{3} ( 9 x )}{3} - (- 9 x + tan (9 x))) : \frac{1}{27} tan^{3} (9 x) + x - \frac{1}{9} tan (9 x)

= \frac{1}{27} tan^{3} (9 x) + x - \frac{1}{9} tan (9 x)

Add a constant to the solution

= \frac{1}{27} tan^{3} (9 x) + x - \frac{1}{9} tan (9 x) + C

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

f (x) = \frac{x}{1 + x ^{2}}, (2, 0.4)

\int_{0}^{2} (x + 2) - x^{2} d x

\frac{d}{d x} ((1 - sin (x))^{- \frac{1}{2}})

What is the integral of tan^4(9x) ?
The integral of tan^4(9x) is 1/27 tan^3(9x)+x-1/9 tan(9x)+C