What is the integral of sec(1-x)tan(1-x) ?

The integral of sec(1-x)tan(1-x) is -sec(1-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(1-x)tan(1-x)

\int sec (1 - x) tan (1 - x) d x

- sec (1 - x) + C

Solution steps

\int sec (1 - x) tan (1 - x) d x

Apply u-substitution

: \int - sec (u) tan (u) d u

= \int - sec (u) tan (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int sec (u) tan (u) d u

Apply u-substitution

: \int 1 d v

= - \int 1 d v

Integral of a constant:

\int a d x = a x

= - 1 \cdot v

Substitute back

= - 1 \cdot sec (1 - x)

Simplify

= - sec (1 - x)

Add a constant to the solution

= - sec (1 - x) + C

\int (- \frac{4}{x ^{3}} - \frac{8}{x ^{5}}) d x

\int \frac{x}{x + x ^{\frac{4}{5}}} d x

\int_{3}^{6} 8 d x

\int - 5 π sin (π x) d x

\int \frac{1}{( 1 + x ) ^{3}} d x

What is the integral of sec(1-x)tan(1-x) ?
The integral of sec(1-x)tan(1-x) is -sec(1-x)+C