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

The integral of (cos(x))/(1+tan(x)) is

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

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

\frac{1}{2} (2 (\frac{1}{2} ln \frac{tan ( \frac{x}{2} ) - 1}{2} + 1 - \frac{1}{2} ln \frac{tan ( \frac{x}{2} ) - 1}{2} - 1) + sin (x) + cos (x)) + C

Solution steps

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

Rewrite using trig identities

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

\frac{1 + cos ( 2 x )}{cos ( x ) + sin ( x )} : \frac{1}{cos ( x ) + sin ( x )} + \frac{cos ( 2 x )}{cos ( x ) + sin ( x )}

Apply the Sum Rule:

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

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

\int \frac{1}{cos ( x ) + sin ( x )} d x = 2 (\frac{1}{2} ln \frac{tan ( \frac{x}{2} ) - 1}{2} + 1 - \frac{1}{2} ln \frac{tan ( \frac{x}{2} ) - 1}{2} - 1)

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

= \frac{1}{2} (2 (\frac{1}{2} ln \frac{tan ( \frac{x}{2} ) - 1}{2} + 1 - \frac{1}{2} ln \frac{tan ( \frac{x}{2} ) - 1}{2} - 1) + sin (x) + cos (x))

Add a constant to the solution

= \frac{1}{2} (2 (\frac{1}{2} ln \frac{tan ( \frac{x}{2} ) - 1}{2} + 1 - \frac{1}{2} ln \frac{tan ( \frac{x}{2} ) - 1}{2} - 1) + sin (x) + cos (x)) + C

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