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

The integral of tan^5(4x)sec^3(4x) is (sec^7(4x))/(28)-(sec^5(4x))/(10)+(sec^3(4x))/(12)+C

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

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

\frac{sec ^{7} ( 4 x )}{28} - \frac{sec ^{5} ( 4 x )}{10} + \frac{sec ^{3} ( 4 x )}{12} + C

Solution steps

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

Rewrite using trig identities

= \int (- 1 + sec^{2} (4 x))^{2} tan (4 x) sec^{3} (4 x) d x

Apply u-substitution

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

Expand

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

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

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

\int \frac{u ^{4}}{2} d u = \frac{u ^{5}}{10}

\int \frac{u ^{2}}{4} d u = \frac{u ^{3}}{12}

= \frac{u ^{7}}{28} - \frac{u ^{5}}{10} + \frac{u ^{3}}{12}

Substitute back

u = sec (4 x)

= \frac{sec ^{7} ( 4 x )}{28} - \frac{sec ^{5} ( 4 x )}{10} + \frac{sec ^{3} ( 4 x )}{12}

Add a constant to the solution

= \frac{sec ^{7} ( 4 x )}{28} - \frac{sec ^{5} ( 4 x )}{10} + \frac{sec ^{3} ( 4 x )}{12} + C

What is the integral of tan^5(4x)sec^3(4x) ?
The integral of tan^5(4x)sec^3(4x) is (sec^7(4x))/(28)-(sec^5(4x))/(10)+(sec^3(4x))/(12)+C