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

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

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

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

- 2 cot (2 t) + C

Solution steps

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

Apply u-substitution

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

\frac{1 + u ^{2}}{u ^{2}} = \frac{1}{u ^{2}} + 1

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

Apply exponent rule:

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

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

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

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

\int 1 d u = u

= - \frac{1}{u} + u

Substitute back

u = tan (t)

= - \frac{1}{tan ( t )} + tan (t)

Simplify

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

= - 2 cot (2 t)

Add a constant to the solution

= - 2 cot (2 t) + C

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