What is the integral of (sec^4(1-x))/(tan^8(1-x)) ?

The integral of (sec^4(1-x))/(tan^8(1-x)) is -(-1/7 cot^7(1-x)-1/5 cot^5(1-x))+C

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integral of (sec^4(1-x))/(tan^8(1-x))

\int \frac{sec ^{4} ( 1 - x )}{tan ^{8} ( 1 - x )} d x

- (- \frac{1}{7} cot^{7} (1 - x) - \frac{1}{5} cot^{5} (1 - x)) + C

Solution steps

\int \frac{sec ^{4} ( 1 - x )}{tan ^{8} ( 1 - x )} d x

Apply u-substitution

: \int - cot^{8} (u) sec^{4} (u) d u

= \int - cot^{8} (u) sec^{4} (u) d u

Take the constant out:

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

= - \int cot^{8} (u) sec^{4} (u) d u

Rewrite using trig identities

= - \int \frac{sec ^{4} ( u )}{tan ^{8} ( u )} d u

Apply u-substitution

: \int \frac{1 + v ^{2}}{v ^{8}} d v

= - \int \frac{1 + v ^{2}}{v ^{8}} d v

Expand

\frac{1 + v ^{2}}{v ^{8}} : \frac{1}{v ^{8}} + \frac{1}{v ^{6}}

= - \int \frac{1}{v ^{8}} + \frac{1}{v ^{6}} 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

= - (\int \frac{1}{v ^{8}} d v + \int \frac{1}{v ^{6}} d v)

\int \frac{1}{v ^{8}} d v = - \frac{1}{7 v ^{7}}

\int \frac{1}{v ^{6}} d v = - \frac{1}{5 v ^{5}}

= - (- \frac{1}{7 v ^{7}} - \frac{1}{5 v ^{5}})

Substitute back

= - (- \frac{1}{7 tan ^{7} ( 1 - x )} - \frac{1}{5 tan ^{5} ( 1 - x )})

Simplify

- (- \frac{1}{7 tan ^{7} ( 1 - x )} - \frac{1}{5 tan ^{5} ( 1 - x )}) : - (- \frac{1}{7} cot^{7} (1 - x) - \frac{1}{5} cot^{5} (1 - x))

= - (- \frac{1}{7} cot^{7} (1 - x) - \frac{1}{5} cot^{5} (1 - x))

Add a constant to the solution

= - (- \frac{1}{7} cot^{7} (1 - x) - \frac{1}{5} cot^{5} (1 - x)) + C

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What is the integral of (sec^4(1-x))/(tan^8(1-x)) ?
The integral of (sec^4(1-x))/(tan^8(1-x)) is -(-1/7 cot^7(1-x)-1/5 cot^5(1-x))+C