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

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

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

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

ln ∣ cos (2 x) ∣ + \frac{1}{2} tan (2 x) + C

Solution steps

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

Apply u-substitution

= \int - tan (u) + \frac{1}{2} sec^{2} (u) d u

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 (u) d u + \int \frac{1}{2} sec^{2} (u) d u

\int tan (u) d u = - ln ∣ cos (u) ∣

\int \frac{1}{2} sec^{2} (u) d u = \frac{1}{2} tan (u)

= - (- ln ∣ cos (u) ∣) + \frac{1}{2} tan (u)

Substitute back

u = 2 x

= - (- ln ∣ cos (2 x) ∣) + \frac{1}{2} tan (2 x)

Simplify

= ln ∣ cos (2 x) ∣ + \frac{1}{2} tan (2 x)

Add a constant to the solution

= ln ∣ cos (2 x) ∣ + \frac{1}{2} tan (2 x) + C

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