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

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

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

\frac{1}{2} tan (2 x - 3) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{2} tan (u)

Substitute back

u = 2 x - 3

= \frac{1}{2} tan (2 x - 3)

Add a constant to the solution

= \frac{1}{2} tan (2 x - 3) + C

\int e^{- s x} cos (a x) d x

\int \frac{x ^{2}}{3 + 4 x - 4 x ^{2}} d x

\int \frac{sec ^{2} ( x )}{9 + tan ^{2} ( x )} d x

\int (x^{2} + 2)^{4} d x

\int \frac{6 ^{x} - 2 ^{x}}{3 ^{x}} d x

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