What is the integral of sec^8(2θ)tan^5(2θ) ?

The integral of sec^8(2θ)tan^5(2θ) is (sec^{12}(2θ))/(24)-(sec^{10}(2θ))/(10)+(sec^8(2θ))/(16)+C

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integral of sec^8(2θ)tan^5(2θ)

\int sec^{8} (2 θ) tan^{5} (2 θ) d θ

\frac{sec ^{12} ( 2 θ )}{24} - \frac{sec ^{10} ( 2 θ )}{10} + \frac{sec ^{8} ( 2 θ )}{16} + C

Solution steps

\int sec^{8} (2 θ) tan^{5} (2 θ) d θ

Rewrite using trig identities

= \int (- 1 + sec^{2} (2 θ))^{2} tan (2 θ) sec^{8} (2 θ) d θ

Apply u-substitution

: \int \frac{u ^{7} ( u ^{2} - 1 ) ^{2}}{2} d u

= \int \frac{u ^{7} ( u ^{2} - 1 ) ^{2}}{2} d u

Expand

\frac{u ^{7} ( u ^{2} - 1 ) ^{2}}{2} : \frac{u ^{11}}{2} - u^{9} + \frac{u ^{7}}{2}

= \int \frac{u ^{11}}{2} - u^{9} + \frac{u ^{7}}{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 ^{11}}{2} d u - \int u^{9} d u + \int \frac{u ^{7}}{2} d u

\int \frac{u ^{11}}{2} d u = \frac{u ^{12}}{24}

\int u^{9} d u = \frac{u ^{10}}{10}

\int \frac{u ^{7}}{2} d u = \frac{u ^{8}}{16}

= \frac{u ^{12}}{24} - \frac{u ^{10}}{10} + \frac{u ^{8}}{16}

Substitute back

u = sec (2 θ)

= \frac{sec ^{12} ( 2 θ )}{24} - \frac{sec ^{10} ( 2 θ )}{10} + \frac{sec ^{8} ( 2 θ )}{16}

Add a constant to the solution

= \frac{sec ^{12} ( 2 θ )}{24} - \frac{sec ^{10} ( 2 θ )}{10} + \frac{sec ^{8} ( 2 θ )}{16} + C

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What is the integral of sec^8(2θ)tan^5(2θ) ?
The integral of sec^8(2θ)tan^5(2θ) is (sec^{12}(2θ))/(24)-(sec^{10}(2θ))/(10)+(sec^8(2θ))/(16)+C