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

The integral of tan^5(2x)sec^3(2x) is (sec^7(2x)}{14}-\frac{sec^5(2x))/5+(sec^3(2x))/6+C

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

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

\frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{5} + \frac{sec ^{3} ( 2 x )}{6} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

\int \frac{u ^{6}}{2} d u = \frac{u ^{7}}{14}

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

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

= \frac{u ^{7}}{14} - \frac{u ^{5}}{5} + \frac{u ^{3}}{6}

Substitute back

u = sec (2 x)

= \frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{5} + \frac{sec ^{3} ( 2 x )}{6}

Add a constant to the solution

= \frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{5} + \frac{sec ^{3} ( 2 x )}{6} + C

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What is the integral of tan^5(2x)sec^3(2x) ?
The integral of tan^5(2x)sec^3(2x) is (sec^7(2x)}{14}-\frac{sec^5(2x))/5+(sec^3(2x))/6+C