What is the integral of tan^5(4x) ?

The integral of tan^5(4x) is 1/4 ln|sec(4x)|-(sec^2(4x))/4+(sec^4(4x))/(16)+C

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integral of tan^5(4x)

\int tan^{5} (4 x) d x

\frac{1}{4} ln ∣ sec (4 x) ∣ - \frac{sec ^{2} ( 4 x )}{4} + \frac{sec ^{4} ( 4 x )}{16} + C

Solution steps

\int tan^{5} (4 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (4 x))^{2} tan (4 x) d x

Apply u-substitution

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

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

Expand

\frac{( - 1 + u ^{2} ) ^{2}}{4 u} : \frac{1}{4 u} - \frac{u}{2} + \frac{u ^{3}}{4}

= \int \frac{1}{4 u} - \frac{u}{2} + \frac{u ^{3}}{4} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{4 u} d u - \int \frac{u}{2} d u + \int \frac{u ^{3}}{4} d u

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

\int \frac{u}{2} d u = \frac{u ^{2}}{4}

\int \frac{u ^{3}}{4} d u = \frac{u ^{4}}{16}

= \frac{1}{4} ln ∣ u ∣ - \frac{u ^{2}}{4} + \frac{u ^{4}}{16}

Substitute back

u = sec (4 x)

= \frac{1}{4} ln ∣ sec (4 x) ∣ - \frac{sec ^{2} ( 4 x )}{4} + \frac{sec ^{4} ( 4 x )}{16}

Add a constant to the solution

= \frac{1}{4} ln ∣ sec (4 x) ∣ - \frac{sec ^{2} ( 4 x )}{4} + \frac{sec ^{4} ( 4 x )}{16} + C

\int \frac{x}{3 x ^{2} + 8 x - 3} d x

x^{^{''}} + 3 x^{^{'}} + 2 x = 0

x + 4

\frac{d y}{d x} + \frac{6 y}{x} = x

x \to 4 lim (arctan (\frac{x ^{2} - 16}{7 x ^{2} - 28 x}))

What is the integral of tan^5(4x) ?
The integral of tan^5(4x) is 1/4 ln|sec(4x)|-(sec^2(4x))/4+(sec^4(4x))/(16)+C