I'm trying to evaluate the following integral:$$\int \frac{{\sin(x)}}{{\sin(x) – \cos(x)}}\,dx$$ Here's my approach so far:
$$$$I've multiplied the numerator and denominator by $-csc^3(x)$, resulting in $$\int -\frac{{\csc^2(x)}}{{\cot(x)\csc^2(x) – \csc^2(x)}}\,dx$$
Next I substituted $u = cotx $, giving me $$\frac{{du}}{{dx}} = -csc^2(x), du = -csc^2(x)dx$$ therefore in $\int -\frac{{\csc^2(x)}}{{\cot(x)\csc^2(x) – \csc^2(x)}}\,dx$ by replacing $cotx$ with $u$, I obtained, $$\int \frac{-1}{{u\left(u^2+1\right) – \left(u^2+1\right)}}\,du$$ $$\int (\frac{1}{{4\left(u+1\right)}} – \frac{1}{{4\left(u-1\right)}} – \frac{1}{{2\left(u-1\right)^2}})\,du$$
I would greatly appreciate any insights or techniques that could help me make progress with this integral. Are there specific strategies or mathematical tools that I should consider? Are there any useful properties or identities related to this type of expression?
Thank you for your attention and assistance.
Best Answer
Alternatively, you can proceed as follows:
\begin{align*} \int\frac{\sin(x)}{\sin(x) - \cos(x)}\mathrm{d}x & = \frac{1}{2}\int\frac{2\sin(x)}{\sin(x) - \cos(x)}\mathrm{d}x\\\\ & = \frac{1}{2}\int\frac{(\sin(x) - \cos(x)) + (\sin(x) + \cos(x))}{\sin(x) - \cos(x)}\mathrm{d}x\\\\ & = \frac{x}{2} + \frac{1}{2}\int\frac{(\sin(x) - \cos(x))'}{\sin(x) - \cos(x)}\mathrm{d}x\\\\ & = \frac{x}{2} + \frac{1}{2}\ln\left(\sin(x) - \cos(x)\right) + C \end{align*}