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

The integral of (sec^2(x))/(tan^4(x)) is -1/3 cot^3(x)+C

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = tan (x)

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} + x y = x y^{2}

\int_{a}^{b} \frac{1}{x ^{3}} d x

f (x) 50 - x

f (x) = 6 + \frac{2}{x} + \frac{3}{x ^{2}}, x = x

f (x) = 2 x^{2} - 3 x + 3

What is the integral of (sec^2(x))/(tan^4(x)) ?
The integral of (sec^2(x))/(tan^4(x)) is -1/3 cot^3(x)+C