How do you find the antiderivative of $\frac{sin x}{1 - cos x}$ $sinx/(1-cosx)$ ?

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How do you find the antiderivative of $\frac{sin x}{1 - cos x}$ $sinx/(1-cosx)$ ?

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Answer 1 · 2016-07-05T20:05:27

You can simply do a u-substitution.

Let:

$u = - cos x$ $u = -cosx$
$d u = sin x d x$ $du = sinxdx$

Therefore, the antiderivative/integral is:

$\int \frac{sin x}{1 - cos x} d x$ $color(blue)(int sinx/(1-cosx)dx)$

$= \int \frac{1}{1 + u} d u$ $= int 1/(1+u)du$

$= ln | 1 + u |$ $= ln|1+u|$

$= ln | 1 - cos x | + C$ $= color(blue)(ln|1-cosx| + C)$

We know this works because:

$\frac{d}{d x} [ln | 1 - cos x | + C]$ $color(green)(d/(dx)[ln|1-cosx| + C])$

$= \frac{1}{1 - cos x} \cdot sin x + 0$ $= 1/(1-cosx)*sinx + 0$
(chain rule)

$= \frac{sin x}{1 - cos x}$ $= color(green)(sinx/(1-cosx))$

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