What is the integral of tan^4(8x) ?

The integral of tan^4(8x) is 1/24 tan^3(8x)+x-1/8 tan(8x)+C

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integral of tan^4(8x)

\int tan^{4} (8 x) d x

\frac{1}{24} tan^{3} (8 x) + x - \frac{1}{8} tan (8 x) + C

Solution steps

\int tan^{4} (8 x) d x

Apply u-substitution

: \int tan^{4} (u) \frac{1}{8} d u

= \int tan^{4} (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int tan^{4} (u) d u

Apply Integral Reduction

= \frac{1}{8} (\frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u)

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

= \frac{1}{8} (\frac{tan ^{3} ( u )}{3} - (- u + tan (u)))

Substitute back

u = 8 x

= \frac{1}{8} (\frac{tan ^{3} ( 8 x )}{3} - (- 8 x + tan (8 x)))

Simplify

\frac{1}{8} (\frac{tan ^{3} ( 8 x )}{3} - (- 8 x + tan (8 x))) : \frac{1}{24} tan^{3} (8 x) + x - \frac{1}{8} tan (8 x)

= \frac{1}{24} tan^{3} (8 x) + x - \frac{1}{8} tan (8 x)

Add a constant to the solution

= \frac{1}{24} tan^{3} (8 x) + x - \frac{1}{8} tan (8 x) + C

\int \frac{1}{v ^{2} + 1} d v

y = 1 - x - x^{2}

\int \frac{x ^{4}}{4} + cosh (x) + C d x

y = e^{2} + 2 x

x \to 0 lim (\frac{e ^{x} - 1}{15 x})

What is the integral of tan^4(8x) ?
The integral of tan^4(8x) is 1/24 tan^3(8x)+x-1/8 tan(8x)+C