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

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

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integral of (1-sin(x))/(1+cos(x))

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

- ln (sec^{2} (\frac{x}{2})) + tan (\frac{x}{2}) + C

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

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

Substitute back

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

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

Simplify

arctan (tan (\frac{x}{2})) - arctan (tan (\frac{x}{2})) + tan (\frac{x}{2}) - ln 1 + tan^{2} (\frac{x}{2}) : - ln (sec^{2} (\frac{x}{2})) + tan (\frac{x}{2})

= - ln (sec^{2} (\frac{x}{2})) + tan (\frac{x}{2})

Add a constant to the solution

= - ln (sec^{2} (\frac{x}{2})) + tan (\frac{x}{2}) + C

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