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

The integral of (sec^4(x))/(tan^4(x)) is -1/3 cot^3(x)-cot(x)+C

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

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

- \frac{1}{3} cot^{3} (x) - cot (x) + C

Solution steps

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

Rewrite using trig identities

= \int (1 + tan^{2} (x)) sec^{2} (x) \frac{1}{tan ^{4} ( x )} d x

Apply u-substitution

: \int \frac{1 + u ^{2}}{u ^{4}} d u

= \int \frac{1 + u ^{2}}{u ^{4}} d u

Expand

\frac{1 + u ^{2}}{u ^{4}} : \frac{1}{u ^{4}} + \frac{1}{u ^{2}}

= \int \frac{1}{u ^{4}} + \frac{1}{u ^{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{1}{u ^{4}} d u + \int \frac{1}{u ^{2}} d u

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

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

= - \frac{1}{3 u ^{3}} - \frac{1}{u}

Substitute back

u = tan (x)

= - \frac{1}{3 tan ^{3} ( x )} - \frac{1}{tan ( x )}

Simplify

- \frac{1}{3 tan ^{3} ( x )} - \frac{1}{tan ( x )} : - \frac{1}{3} cot^{3} (x) - cot (x)

= - \frac{1}{3} cot^{3} (x) - cot (x)

Add a constant to the solution

= - \frac{1}{3} cot^{3} (x) - cot (x) + C

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What is the integral of (sec^4(x))/(tan^4(x)) ?
The integral of (sec^4(x))/(tan^4(x)) is -1/3 cot^3(x)-cot(x)+C