What is the integral of 1/(1+tan(β)) ?

The integral of 1/(1+tan(β)) is 1/2 (-1/2 ln(sec^2(β))+arctan(tan(β)))+1/2 ln|tan(β)+1|+C

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integral of 1/(1+tan(β))

\int \frac{1}{1 + tan ( β )} d β

\frac{1}{2} (- \frac{1}{2} ln (sec^{2} (β)) + arctan (tan (β))) + \frac{1}{2} ln ∣ tan (β) + 1 ∣ + C

Solution steps

\int \frac{1}{1 + tan ( β )} d β

Apply u-substitution

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

Take the partial fraction of

\frac{1}{( 1 + u ^{2} ) ( 1 + u )} : \frac{- u + 1}{2 ( u ^{2} + 1 )} + \frac{1}{2 ( u + 1 )}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{1}{2 ( u + 1 )} d u = \frac{1}{2} ln ∣ u + 1 ∣

= \frac{1}{2} (- \frac{1}{2} ln u^{2} + 1 + arctan (u)) + \frac{1}{2} ln ∣ u + 1 ∣

Substitute back

u = tan (β)

= \frac{1}{2} (- \frac{1}{2} ln tan^{2} (β) + 1 + arctan (tan (β))) + \frac{1}{2} ln ∣ tan (β) + 1 ∣

\frac{1}{2} ln tan^{2} (β) + 1 = \frac{1}{2} ln (sec^{2} (β))

= \frac{1}{2} (- \frac{1}{2} ln (sec^{2} (β)) + arctan (tan (β))) + \frac{1}{2} ln ∣ tan (β) + 1 ∣

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{2} ln (sec^{2} (β)) + arctan (tan (β))) + \frac{1}{2} ln ∣ tan (β) + 1 ∣ + C

What is the integral of 1/(1+tan(β)) ?
The integral of 1/(1+tan(β)) is 1/2 (-1/2 ln(sec^2(β))+arctan(tan(β)))+1/2 ln|tan(β)+1|+C