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

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

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integral of sec^{13}(2θ)tan^3(2θ)

\int sec^{13} (2 θ) tan^{3} (2 θ) d θ

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

Solution steps

\int sec^{13} (2 θ) tan^{3} (2 θ) d θ

Rewrite using trig identities

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

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 θ)

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

Add a constant to the solution

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

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