What is the integral of sec^2(3x-1) ?

The integral of sec^2(3x-1) is -1/3 tan(1-3x)+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(3x-1)

\int sec^{2} (3 x - 1) d x

- \frac{1}{3} tan (1 - 3 x) + C

Solution steps

\int sec^{2} (3 x - 1) d x

Factor

3 x - 1 : - (- 3 x + 1)

= \int sec^{2} (- (- 3 x + 1)) d x

Simplify

sec^{2} (- (- 3 x + 1)) : sec^{2} (1 - 3 x)

= \int sec^{2} (1 - 3 x) d x

Apply u-substitution

: \int - \frac{1}{3} sec^{2} (u) d u

= \int - \frac{1}{3} sec^{2} (u) d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= - \frac{1}{3} tan (u)

Substitute back

u = 1 - 3 x

= - \frac{1}{3} tan (1 - 3 x)

Add a constant to the solution

= - \frac{1}{3} tan (1 - 3 x) + C

\int x \cdot 3^{x^{2}} d x

\int (1 - y)^{4} d y

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

\int 13 sin^{3} (x) cos^{2} (x) d x

What is the integral of sec^2(3x-1) ?
The integral of sec^2(3x-1) is -1/3 tan(1-3x)+C