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

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

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

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

arctan (sin (x)) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 1} d u

= \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= arctan (u)

Substitute back

u = sin (x)

= arctan (sin (x))

Add a constant to the solution

= arctan (sin (x)) + C

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