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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

ln ∣ 1 + sin (t) ∣ + C

Solution steps

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

Rewrite using trig identities

= \int \frac{cos ( t )}{1 + sin ( 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 + sin (t)

= ln ∣ 1 + sin (t) ∣

Add a constant to the solution

= ln ∣ 1 + sin (t) ∣ + C

n = 1 \sum \infty \frac{1}{5 n + 1}

\int (x x + 1) d x

0.9 x^{2} + 6, x, (- 4, 4)

\frac{d y}{d x} = e^{2 x} + 4 y

\frac{d}{d x} (3 x + 2 x^{2})

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