What is the integral of (sec^4(2x))/(tan^9(2x)) ?

The integral of (sec^4(2x))/(tan^9(2x)) is 1/2 (-1/8 cot^8(2x)-1/6 cot^6(2x))+C

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integral of (sec^4(2x))/(tan^9(2x))

\int \frac{sec ^{4} ( 2 x )}{tan ^{9} ( 2 x )} d x

\frac{1}{2} (- \frac{1}{8} cot^{8} (2 x) - \frac{1}{6} cot^{6} (2 x)) + C

Solution steps

\int \frac{sec ^{4} ( 2 x )}{tan ^{9} ( 2 x )} d x

Apply u-substitution

: \int \frac{sec ^{4} ( u )}{2 tan ^{9} ( u )} d u

= \int \frac{sec ^{4} ( u )}{2 tan ^{9} ( u )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{sec ^{4} ( u )}{tan ^{9} ( u )} d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) \frac{1}{tan ^{9} ( u )} d u

Apply u-substitution

: \int \frac{1 + v ^{2}}{v ^{9}} d v

= \frac{1}{2} \cdot \int \frac{1 + v ^{2}}{v ^{9}} d v

Expand

\frac{1 + v ^{2}}{v ^{9}} : \frac{1}{v ^{9}} + \frac{1}{v ^{7}}

= \frac{1}{2} \cdot \int \frac{1}{v ^{9}} + \frac{1}{v ^{7}} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{1}{v ^{9}} d v + \int \frac{1}{v ^{7}} d v)

\int \frac{1}{v ^{9}} d v = - \frac{1}{8 v ^{8}}

\int \frac{1}{v ^{7}} d v = - \frac{1}{6 v ^{6}}

= \frac{1}{2} (- \frac{1}{8 v ^{8}} - \frac{1}{6 v ^{6}})

Substitute back

= \frac{1}{2} (- \frac{1}{8 tan ^{8} ( 2 x )} - \frac{1}{6 tan ^{6} ( 2 x )})

Simplify

\frac{1}{2} (- \frac{1}{8 tan ^{8} ( 2 x )} - \frac{1}{6 tan ^{6} ( 2 x )}) : \frac{1}{2} (- \frac{1}{8} cot^{8} (2 x) - \frac{1}{6} cot^{6} (2 x))

= \frac{1}{2} (- \frac{1}{8} cot^{8} (2 x) - \frac{1}{6} cot^{6} (2 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{8} cot^{8} (2 x) - \frac{1}{6} cot^{6} (2 x)) + C

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What is the integral of (sec^4(2x))/(tan^9(2x)) ?
The integral of (sec^4(2x))/(tan^9(2x)) is 1/2 (-1/8 cot^8(2x)-1/6 cot^6(2x))+C