What is the integral of tan^4(2x) ?

The integral of tan^4(2x) is 1/6 tan^3(2x)+x-1/2 tan(2x)+C

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integral of tan^4(2x)

\int tan^{4} (2 x) d x

\frac{1}{6} tan^{3} (2 x) + x - \frac{1}{2} tan (2 x) + C

Solution steps

\int tan^{4} (2 x) d x

Apply u-substitution

: \int tan^{4} (u) \frac{1}{2} d u

= \int tan^{4} (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int tan^{4} (u) d u

Apply Integral Reduction

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

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

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

Substitute back

u = 2 x

= \frac{1}{2} (\frac{tan ^{3} ( 2 x )}{3} - (- 2 x + tan (2 x)))

Simplify

\frac{1}{2} (\frac{tan ^{3} ( 2 x )}{3} - (- 2 x + tan (2 x))) : \frac{1}{6} tan^{3} (2 x) + x - \frac{1}{2} tan (2 x)

= \frac{1}{6} tan^{3} (2 x) + x - \frac{1}{2} tan (2 x)

Add a constant to the solution

= \frac{1}{6} tan^{3} (2 x) + x - \frac{1}{2} tan (2 x) + C

t^{5} y^{4} - t^{3} y^{^{'}} + 6 y = 0

y^{^{'}} = \frac{3 ( e ^{x} + 1 )}{2 y}

\frac{d}{dd} (\frac{a x + b}{c x + d})

\frac{d}{d x} (- \frac{1}{2} x^{4} cos (x))

What is the integral of tan^4(2x) ?
The integral of tan^4(2x) is 1/6 tan^3(2x)+x-1/2 tan(2x)+C