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

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

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

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

arctan (sin (t)) + C

Solution steps

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

Rewrite using trig identities

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

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 (t)

= arctan (sin (t))

Add a constant to the solution

= arctan (sin (t)) + C

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