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)

integral

sec^{2} (2 x - 1)

\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 = [3 x + 2] [2 x - 5]

\frac{1}{( k + e ^{- t} )}

e^{\frac{y ^{2}}{2}}

x (x^{2})

\frac{1}{( 1 - 2 x ^{2} )}

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