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

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

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integral of (sec(2x)-1)/(sec(2x)+1)

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

- x + tan (x) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1 - cos ( 2 x )}{1 + cos ( 2 x )} d x

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

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

Rewrite using trig identities

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

\int 1 d x = x

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

= - x + tan (x)

Add a constant to the solution

= - x + tan (x) + C

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\int \frac{x ^{3} + x}{x ^{4} - 2 x ^{2} + 1} d x

\int \frac{x}{2 x ^{2} - 3 x - 2} d x

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