What is the integral of (sec^4(θ))/(tan^5(θ)) ?

The integral of (sec^4(θ))/(tan^5(θ)) is -1/4 cot^4(θ)-1/2 cot^2(θ)+C

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integral of (sec^4(θ))/(tan^5(θ))

\int \frac{sec ^{4} ( θ )}{tan ^{5} ( θ )} d θ

- \frac{1}{4} cot^{4} (θ) - \frac{1}{2} cot^{2} (θ) + C

Solution steps

\int \frac{sec ^{4} ( θ )}{tan ^{5} ( θ )} d θ

Rewrite using trig identities

= \int (1 + tan^{2} (θ)) sec^{2} (θ) \frac{1}{tan ^{5} ( θ )} d θ

Apply u-substitution

: \int \frac{1 + u ^{2}}{u ^{5}} d u

= \int \frac{1 + u ^{2}}{u ^{5}} d u

Expand

\frac{1 + u ^{2}}{u ^{5}} : \frac{1}{u ^{5}} + \frac{1}{u ^{3}}

= \int \frac{1}{u ^{5}} + \frac{1}{u ^{3}} 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{1}{u ^{5}} d u + \int \frac{1}{u ^{3}} d u

\int \frac{1}{u ^{5}} d u = - \frac{1}{4 u ^{4}}

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

= - \frac{1}{4 u ^{4}} - \frac{1}{2 u ^{2}}

Substitute back

u = tan (θ)

= - \frac{1}{4 tan ^{4} ( θ )} - \frac{1}{2 tan ^{2} ( θ )}

Simplify

- \frac{1}{4 tan ^{4} ( θ )} - \frac{1}{2 tan ^{2} ( θ )} : - \frac{1}{4} cot^{4} (θ) - \frac{1}{2} cot^{2} (θ)

= - \frac{1}{4} cot^{4} (θ) - \frac{1}{2} cot^{2} (θ)

Add a constant to the solution

= - \frac{1}{4} cot^{4} (θ) - \frac{1}{2} cot^{2} (θ) + C

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What is the integral of (sec^4(θ))/(tan^5(θ)) ?
The integral of (sec^4(θ))/(tan^5(θ)) is -1/4 cot^4(θ)-1/2 cot^2(θ)+C