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

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

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integral of sec^2(2x-1)

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

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

Solution steps

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

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

Add a constant to the solution

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

y (x - y - 2) d x + x (y - x + 4) d y = 0

e^{2} y \frac{d y}{d x} = e^{- y} + e^{- 2 x - y}

(- 4, 7), (- 6, - 4)

\frac{d}{d x} (ln (cos (sin (e^{x^{3}}))))

y^{^{''''}} + 13 y^{^{''}} + 36 y = 0

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