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

integral

sec^{2} (2 x) - 1

\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

\frac{t ^{3}}{t ^{2} + 1}

\frac{1}{( x ^{2} - 6 x + 5 )}

\frac{5}{x + 1}

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

\frac{( 1 - x ^{2} )}{( e ^{2} x )}

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