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

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

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integral of sec^2(2x)-1

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

\frac{1}{2} tan (2 x) - x + C

Solution steps

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

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

\int 1 d x = x

= \frac{1}{2} tan (2 x) - x

Add a constant to the solution

= \frac{1}{2} tan (2 x) - x + C

\int tan^{2} (x) sec^{10} (x) d x

x \to 0 + lim (\frac{tan ( x )}{x})

y^{^{''''}} + 50 y^{^{''}} + 625 y = 0

x \to 3 lim (2 x^{3} - 10 x - 12)

x \to 2 lim (4 + 5 (- 2))

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