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

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

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integral of sec^2(x-1)

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

- tan (1 - x) + C

Solution steps

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

Factor

x - 1 : - (- x + 1)

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

Simplify

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= - tan (u)

Substitute back

u = 1 - x

= - tan (1 - x)

Add a constant to the solution

= - tan (1 - x) + C

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