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

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

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integral of (tan(x)-1)^2

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

tan (x) + 2 ln ∣ cos (x) ∣ + C

Solution steps

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

Expand

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

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

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

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

\int 1 d x = x

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

Simplify

= tan (x) + 2 ln ∣ cos (x) ∣

Add a constant to the solution

= tan (x) + 2 ln ∣ cos (x) ∣ + C

\frac{d}{d x} (\frac{1 + x ^{2}}{1 + y ^{2}})

x \to 0 lim (\frac{e x}{x})

\int_{π}^{\frac{3 π}{2}} x sin (x) d x

\frac{d y}{d x} = 2 sin (x) (y - 1)

\int (\frac{ln ( x )}{x})^{3} d x

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