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

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

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

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

\frac{1}{2} (ln tan (\frac{x}{2}) + \frac{tan ^{2} ( \frac{x}{2} )}{2}) + C

Solution steps

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

Apply u-substitution

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

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

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 + u ^{2}}{u} d u

Expand

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

= \frac{1}{2} \cdot \int \frac{1}{u} + u 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

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int u d u = \frac{u ^{2}}{2}

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

Substitute back

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

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

Add a constant to the solution

= \frac{1}{2} (ln tan (\frac{x}{2}) + \frac{tan ^{2} ( \frac{x}{2} )}{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 1/2 (ln|tan(x/2)|+(tan^2(x/2))/2)+C