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

The integral of tan^4(4x) is 1/12 tan^3(4x)+x-1/4 tan(4x)+C

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integral of tan^4(4x)

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

\frac{1}{12} tan^{3} (4 x) + x - \frac{1}{4} tan (4 x) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

= \frac{1}{4} (\frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u)

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

= \frac{1}{4} (\frac{tan ^{3} ( u )}{3} - (- u + tan (u)))

Substitute back

u = 4 x

= \frac{1}{4} (\frac{tan ^{3} ( 4 x )}{3} - (- 4 x + tan (4 x)))

Simplify

\frac{1}{4} (\frac{tan ^{3} ( 4 x )}{3} - (- 4 x + tan (4 x))) : \frac{1}{12} tan^{3} (4 x) + x - \frac{1}{4} tan (4 x)

= \frac{1}{12} tan^{3} (4 x) + x - \frac{1}{4} tan (4 x)

Add a constant to the solution

= \frac{1}{12} tan^{3} (4 x) + x - \frac{1}{4} tan (4 x) + C

6 x^{l n (x)}

\int_{1}^{9} π (x - 1) d x

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

f (x) = \frac{5}{x}, (5, 1)

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

What is the integral of tan^4(4x) ?
The integral of tan^4(4x) is 1/12 tan^3(4x)+x-1/4 tan(4x)+C