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

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

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integral of tan^2(x-1)

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

- x - tan (1 - x) + C

Solution steps

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

Factor

x - 1 : - (- x + 1)

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

Simplify

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

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

Rewrite using trig identities

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d x = x

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

= - x - tan (1 - x)

Add a constant to the solution

= - x - tan (1 - x) + C

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