MCQ
$\int_{}^{} {\frac{{\sin x}}{{\sin x - \cos x}}} \;dx = $
- A$\frac{1}{2}\log (\sin x - \cos x) + x + c$
- ✓$\frac{1}{2}[\log (\sin x - \cos x) + x] + c$
- C$\frac{1}{2}\log (\cos x - \sin x) + x + c$
- D$\frac{1}{2}[\log (\cos x - \sin x) + x] + c$
✓
Answer
Correct option: B.
$\frac{1}{2}[\log (\sin x - \cos x) + x] + c$
b
(b)$\int_{}^{} {\frac{{\sin x\,dx}}{{\sin x - \cos x}}} = \frac{1}{2}\int_{}^{} {\frac{{2\sin x}}{{\sin x - \cos x}}\,dx} $
$ = \frac{1}{2}\int_{}^{} {\frac{{(\sin x - \cos x + \sin x + \cos x)}}{{\sin x - \cos x}}\,dx} $
$ = \frac{1}{2}\int_{}^{} {\left( {1 + \frac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right)\,dx} $
(b)$\int_{}^{} {\frac{{\sin x\,dx}}{{\sin x - \cos x}}} = \frac{1}{2}\int_{}^{} {\frac{{2\sin x}}{{\sin x - \cos x}}\,dx} $
$ = \frac{1}{2}\int_{}^{} {\frac{{(\sin x - \cos x + \sin x + \cos x)}}{{\sin x - \cos x}}\,dx} $
$ = \frac{1}{2}\int_{}^{} {\left( {1 + \frac{{\sin x + \cos x}}{{\sin x - \cos x}}} \right)\,dx} $
$= \frac{1}{2}[x + \log (\sin x - \cos x)] + c$.
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