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

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

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integral of (tan(θ)-1)^2

\int (tan (θ) - 1)^{2} d θ

tan (θ) + 2 ln ∣ cos (θ) ∣ + C

Solution steps

\int (tan (θ) - 1)^{2} d θ

Expand

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

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

\int tan^{2} (θ) d θ = - θ + tan (θ)

\int 2 tan (θ) d θ = - 2 ln ∣ cos (θ) ∣

\int 1 d θ = θ

= - θ + tan (θ) - (- 2 ln ∣ cos (θ) ∣) + θ

Simplify

= tan (θ) + 2 ln ∣ cos (θ) ∣

Add a constant to the solution

= tan (θ) + 2 ln ∣ cos (θ) ∣ + C

\frac{d}{d x} (x^{\frac{8}{3}})

\int_{1}^{\infty} e^{- 7 x} d x

\frac{d}{d x} (a cos (x - \frac{π}{2}) + b)

\int e^{- t (s + 5)} d t

e^{2 x}, 2

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