What is the integral of (sec^4(t))/(tan^3(t)) ?

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

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integral of (sec^4(t))/(tan^3(t))

\int \frac{sec ^{4} ( t )}{tan ^{3} ( t )} d t

- \frac{1}{2} cot^{2} (t) + ln ∣ tan (t) ∣ + C

Solution steps

\int \frac{sec ^{4} ( t )}{tan ^{3} ( t )} d t

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{sec ^{4} ( t )}{tan ^{3} ( t )} = (sec^{4} (t)) tan^{- 3} (t)

= \int sec^{4} (t) tan^{- 3} (t) d t

Rewrite using trig identities

= \int (1 + tan^{2} (t)) sec^{2} (t) tan^{- 3} (t) d t

Apply u-substitution

: \int \frac{1 + u ^{2}}{u ^{3}} d u

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

Expand

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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

= - \frac{1}{2 u ^{2}} + ln ∣ u ∣

Substitute back

u = tan (t)

= - \frac{1}{2 tan ^{2} ( t )} + ln ∣ tan (t) ∣

Simplify

- \frac{1}{2 tan ^{2} ( t )} + ln ∣ tan (t) ∣ : - \frac{1}{2} cot^{2} (t) + ln ∣ tan (t) ∣

= - \frac{1}{2} cot^{2} (t) + ln ∣ tan (t) ∣

Add a constant to the solution

= - \frac{1}{2} cot^{2} (t) + ln ∣ tan (t) ∣ + C

\frac{x ^{2} + 3}{x - 1}

x \to 0 lim ((e^{x} + 3 x)^{\frac{1}{x}})

\int cos (n) t d t

(- 3 + 18 x - 45 x^{2}) (x + 4)

\int (\frac{1}{e ^{x}}) d x

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