What is the integral of 2tan^5(4x) ?

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

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integral of 2tan^5(4x)

\int 2 tan^{5} (4 x) d x

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

Solution steps

\int 2 tan^{5} (4 x) d x

Take the constant out:

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

= 2 \cdot \int tan^{5} (4 x) d x

Rewrite using trig identities

= 2 \cdot \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

= 2 \cdot \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}

= 2 \cdot \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

= 2 (\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}

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

Substitute back

u = sec (4 x)

= 2 (\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

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

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\frac{2}{( 4 s ^{2} - 8 s )}

\frac{d}{d x} (sin (x) + arcsin (x))

\int \frac{x}{sin ^{2} ( x ^{2} )} d x

12 x^{3} - 3 y^{2} x^{4} + 1 (\frac{d y}{d x}) = 0

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