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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

ln ∣ 1 + tan (x) ∣ + C

Solution steps

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

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 (x)

= ln ∣ 1 + tan (x) ∣

Add a constant to the solution

= ln ∣ 1 + tan (x) ∣ + C

n = 1 \sum \infty \frac{3 n}{4 n + 1}

\frac{( x + 2 ) ^{(2)} ( d y )}{( d x )} = 5 - 8 y - 4 x y

\int \frac{( y - 2 )}{y + 3} d y

\int_{0}^{0.6} 2 x e^{- x} d x

y^{^{'}} = 2 y

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