What is the integral of 1/(sec(x)(1-sin^2(x))) ?

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

\int \frac{1}{sec ( x ) ( 1 - sin ^{2} ( x ) )} d x

ln ∣ tan (x) + sec (x) ∣ + C

Solution steps

\int \frac{1}{sec ( x ) ( 1 - sin ^{2} ( x ) )} d x

Simplify

\frac{1}{sec ( x ) ( 1 - sin ^{2} ( x ) )} : sec (x)

= \int sec (x) d x

Use the common integral:

\int sec (x) d x = ln ∣ tan (x) + sec (x) ∣

= ln ∣ tan (x) + sec (x) ∣

Add a constant to the solution

= ln ∣ tan (x) + sec (x) ∣ + C

\int x^{2} e^{\frac{x}{7}} d x

\int \frac{1}{1 + 4 x ^{2} + 1} d x

\int \frac{x}{4 9 - x ^{2}} d x

\int \frac{1}{1 + e ^{11 x}} d x

\int \frac{x + 3}{x ^{2} + 2 x + 10} d x

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