What is the integral of (sec^2(t))/(1+tan(t)) ?

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

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integral of (sec^2(t))/(1+tan(t))

\int \frac{sec ^{2} ( t )}{1 + tan ( t )} d t

ln ∣ 1 + tan (t) ∣ + C

Solution steps

\int \frac{sec ^{2} ( t )}{1 + tan ( t )} d t

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = 1 + tan (t)

= ln ∣ 1 + tan (t) ∣

Add a constant to the solution

= ln ∣ 1 + tan (t) ∣ + C

\int (\frac{2 y ^{\frac{1}{2}}}{y ^{3}}) d^{0} x

\frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} - 4})

(1 + ln (x) + \frac{y}{x}) d x + (- 1 + ln (x)) d y = 0

y = e^{3 x} + 10 e^{2 x}

x \to 0 lim (\frac{2 e ^{x} - 2}{3 x})

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