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

The integral of sec(4x)tan^5(4x) is 1/4 sec(4x)-(sec^3(4x))/6+(sec^5(4x))/(20)+C

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

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

\frac{1}{4} sec (4 x) - \frac{sec ^{3} ( 4 x )}{6} + \frac{sec ^{5} ( 4 x )}{20} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

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

\int \frac{u ^{2}}{2} d u = \frac{u ^{3}}{6}

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

= \frac{1}{4} u - \frac{u ^{3}}{6} + \frac{u ^{5}}{20}

Substitute back

u = sec (4 x)

= \frac{1}{4} sec (4 x) - \frac{sec ^{3} ( 4 x )}{6} + \frac{sec ^{5} ( 4 x )}{20}

Add a constant to the solution

= \frac{1}{4} sec (4 x) - \frac{sec ^{3} ( 4 x )}{6} + \frac{sec ^{5} ( 4 x )}{20} + C

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What is the integral of sec(4x)tan^5(4x) ?
The integral of sec(4x)tan^5(4x) is 1/4 sec(4x)-(sec^3(4x))/6+(sec^5(4x))/(20)+C