What is the integral of (sec(x)+tan(x))^{-1} ?

The integral of (sec(x)+tan(x))^{-1} is ln|1+sin(x)|+C

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integral of (sec(x)+tan(x))^{-1}

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

ln ∣ 1 + sin (x) ∣ + C

Solution steps

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = 1 + sin (x)

= ln ∣ 1 + sin (x) ∣

Add a constant to the solution

= ln ∣ 1 + sin (x) ∣ + C

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What is the integral of (sec(x)+tan(x))^{-1} ?
The integral of (sec(x)+tan(x))^{-1} is ln|1+sin(x)|+C