Integration

$\begin{aligned} \int \dfrac{\sin x}{1 - \cos x} \ dx & = ?\\ \\ \text{Put t} & = 1 - \cos x\\ \text{Differentiate on both sides}:-\\ dt & = 0 - \left(-\sin x\right) \ dx\\ dt & = \sin x \ dx\\ \\ \text{Substituting in the integral}:-\\ \int \dfrac{1}{t} \ dt & = ln\left(t\right) + \small \color{grey}{constant}\\ \\ & = \color{#3D99F6}{\boxed{\ln\left(1 - \cos x\right) + \small \color{grey}{constant}}} \end{aligned}$

$\begin{aligned} I & = \int \frac {\sin x}{1-\cos x} dx & \small \color{#3D99F6} \text{Since }d (1-\cos x) = \sin x \ dx \\ & = \int \frac {d(1-\cos x)}{1-\cos x} \\ & = \boxed{\ln (1-\cos x) + \color{#3D99F6}C} & \small \color{#3D99F6} \text{where }C \text{ is constant of integration.} \end{aligned}$

I solved this by taking the derivative of each of the answer choices. While you could just take the antiderivative of the integrand, I found this to be easier.

Here are the answer choices and their derivative:

(Cosx)^2 --> 2(cosx)(sinx) ln(1-cosx)+c --> 1/(1-cosx)×(sinx) ln (sinx-cosx)+c --> 1/(sinx-cosx)×(cosx+sinx)

After taking the derivatives (which if you don't know how to do by hand (or head) then you might want to use a calculator) you can clearly see that thensecond choice above is the same. If you were to check this on a calculator, you would find that this is correct.