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

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

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integral of 1/(sec(x)(1+cos(x)))

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

2 arctan (tan (\frac{x}{2})) - tan (\frac{x}{2}) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

\frac{1 - u ^{2}}{1 + u ^{2}} = \frac{2}{1 + u ^{2}} - 1

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

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

\int 1 d u = u

= 2 arctan (u) - u

Substitute back

u = tan (\frac{x}{2})

= 2 arctan (tan (\frac{x}{2})) - tan (\frac{x}{2})

Add a constant to the solution

= 2 arctan (tan (\frac{x}{2})) - tan (\frac{x}{2}) + C

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