What is the integral of tan^2(θ-1) ?

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

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integral of tan^2(θ-1)

\int tan^{2} (θ - 1) d θ

- θ - tan (1 - θ) + C

Solution steps

\int tan^{2} (θ - 1) d θ

Factor

θ - 1 : - (- θ + 1)

= \int tan^{2} (- (- θ + 1)) d θ

Simplify

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

= \int tan^{2} (1 - θ) d θ

Rewrite using trig identities

= \int - 1 + sec^{2} (1 - θ) d θ

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 θ + \int sec^{2} (1 - θ) d θ

\int 1 d θ = θ

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

= - θ - tan (1 - θ)

Add a constant to the solution

= - θ - tan (1 - θ) + C

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