What is the integral of tan^4(11x) ?

The integral of tan^4(11x) is 1/33 tan^3(11x)+x-1/11 tan(11x)+C

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integral of tan^4(11x)

\int tan^{4} (11 x) d x

\frac{1}{33} tan^{3} (11 x) + x - \frac{1}{11} tan (11 x) + C

Solution steps

\int tan^{4} (11 x) d x

Apply u-substitution

: \int tan^{4} (u) \frac{1}{11} d u

= \int tan^{4} (u) \frac{1}{11} d u

Take the constant out:

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

= \frac{1}{11} \cdot \int tan^{4} (u) d u

Apply Integral Reduction

= \frac{1}{11} (\frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u)

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

= \frac{1}{11} (\frac{tan ^{3} ( u )}{3} - (- u + tan (u)))

Substitute back

u = 11 x

= \frac{1}{11} (\frac{tan ^{3} ( 11 x )}{3} - (- 11 x + tan (11 x)))

Simplify

\frac{1}{11} (\frac{tan ^{3} ( 11 x )}{3} - (- 11 x + tan (11 x))) : \frac{1}{33} tan^{3} (11 x) + x - \frac{1}{11} tan (11 x)

= \frac{1}{33} tan^{3} (11 x) + x - \frac{1}{11} tan (11 x)

Add a constant to the solution

= \frac{1}{33} tan^{3} (11 x) + x - \frac{1}{11} tan (11 x) + C

\frac{\partial}{\partial z} (4 sin^{2} (z))

\int (x - \frac{1}{x}) d x

\frac{d}{d x} (e^{2^{x}})

d x + e^{- 5 x} d y = 0

f (x) = 2 ln (6 + x)

What is the integral of tan^4(11x) ?
The integral of tan^4(11x) is 1/33 tan^3(11x)+x-1/11 tan(11x)+C