What is the integral of sec^{13}(2x)tan^3(2x) ?

The integral of sec^{13}(2x)tan^3(2x) is (sec^{15}(2x))/(30)-(sec^{13}(2x))/(26)+C

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integral of sec^{13}(2x)tan^3(2x)

\int sec^{13} (2 x) tan^{3} (2 x) d x

\frac{sec ^{15} ( 2 x )}{30} - \frac{sec ^{13} ( 2 x )}{26} + C

Solution steps

\int sec^{13} (2 x) tan^{3} (2 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (2 x)) tan (2 x) sec^{13} (2 x) d x

Apply u-substitution

: \int \frac{u ^{12} ( u ^{2} - 1 )}{2} d u

= \int \frac{u ^{12} ( u ^{2} - 1 )}{2} d u

Expand

\frac{u ^{12} ( u ^{2} - 1 )}{2} : \frac{u ^{14}}{2} - \frac{u ^{12}}{2}

= \int \frac{u ^{14}}{2} - \frac{u ^{12}}{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 ^{14}}{2} d u - \int \frac{u ^{12}}{2} d u

\int \frac{u ^{14}}{2} d u = \frac{u ^{15}}{30}

\int \frac{u ^{12}}{2} d u = \frac{u ^{13}}{26}

= \frac{u ^{15}}{30} - \frac{u ^{13}}{26}

Substitute back

u = sec (2 x)

= \frac{sec ^{15} ( 2 x )}{30} - \frac{sec ^{13} ( 2 x )}{26}

Add a constant to the solution

= \frac{sec ^{15} ( 2 x )}{30} - \frac{sec ^{13} ( 2 x )}{26} + C

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What is the integral of sec^{13}(2x)tan^3(2x) ?
The integral of sec^{13}(2x)tan^3(2x) is (sec^{15}(2x))/(30)-(sec^{13}(2x))/(26)+C