What is the integral of sec^2(5x-1) ?

The integral of sec^2(5x-1) is -1/5 tan(1-5x)+C

integral of sec^2(5x-1)

\int sec^{2} (5 x - 1) d x

- \frac{1}{5} tan (1 - 5 x) + C

Solution steps

\int sec^{2} (5 x - 1) d x

Factor

5 x - 1 : - (- 5 x + 1)

= \int sec^{2} (- (- 5 x + 1)) d x

Simplify

sec^{2} (- (- 5 x + 1)) : sec^{2} (1 - 5 x)

= \int sec^{2} (1 - 5 x) d x

Apply u-substitution

= \int - \frac{1}{5} 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}{5} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= - \frac{1}{5} tan (u)

Substitute back

u = 1 - 5 x

= - \frac{1}{5} tan (1 - 5 x)

Add a constant to the solution

= - \frac{1}{5} tan (1 - 5 x) + C

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What is the integral of sec^2(5x-1) ?
The integral of sec^2(5x-1) is -1/5 tan(1-5x)+C