What is the integral of 2tan^3(4x)sec^5(4x) ?

The integral of 2tan^3(4x)sec^5(4x) is 2((sec^7(4x))/(28)-(sec^5(4x))/(20))+C

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integral of 2tan^3(4x)sec^5(4x)

\int 2 tan^{3} (4 x) sec^{5} (4 x) d x

2 (\frac{sec ^{7} ( 4 x )}{28} - \frac{sec ^{5} ( 4 x )}{20}) + C

Solution steps

\int 2 tan^{3} (4 x) sec^{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^{3} (4 x) sec^{5} (4 x) d x

Rewrite using trig identities

= 2 \cdot \int (- 1 + sec^{2} (4 x)) tan (4 x) sec^{5} (4 x) d x

Apply u-substitution

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

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

Expand

\frac{u ^{4} ( u ^{2} - 1 )}{4} : \frac{u ^{6}}{4} - \frac{u ^{4}}{4}

= 2 \cdot \int \frac{u ^{6}}{4} - \frac{u ^{4}}{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{u ^{6}}{4} d u - \int \frac{u ^{4}}{4} d u)

\int \frac{u ^{6}}{4} d u = \frac{u ^{7}}{28}

\int \frac{u ^{4}}{4} d u = \frac{u ^{5}}{20}

= 2 (\frac{u ^{7}}{28} - \frac{u ^{5}}{20})

Substitute back

u = sec (4 x)

= 2 (\frac{sec ^{7} ( 4 x )}{28} - \frac{sec ^{5} ( 4 x )}{20})

Add a constant to the solution

= 2 (\frac{sec ^{7} ( 4 x )}{28} - \frac{sec ^{5} ( 4 x )}{20}) + C

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What is the integral of 2tan^3(4x)sec^5(4x) ?
The integral of 2tan^3(4x)sec^5(4x) is 2((sec^7(4x))/(28)-(sec^5(4x))/(20))+C