MCQ
$\int_{}^{} {\frac{{\cos x - \sin x}}{{1 + \sin 2x}}\;dx = } $
- ✓$ - \frac{1}{{\cos x + \sin x}} + c$
- B$\frac{1}{{\cos x + \sin x}} + c$
- C$\frac{1}{{\cos x - \sin x}} + c$
- DNone of these
✓
Answer
Correct option: A.
$ - \frac{1}{{\cos x + \sin x}} + c$
a
(a)$\int_{}^{} {\frac{{\cos x - \sin x}}{{1 + \sin 2x}}\,dx} = \int_{}^{} {\frac{{\cos x - \sin x}}{{{{(\sin x + \cos x)}^2}}}\,dx} $
Now put $\sin x + \cos x = t,$ then the $ \Rightarrow (\cos x - \sin x)\,dx = dt$
(a)$\int_{}^{} {\frac{{\cos x - \sin x}}{{1 + \sin 2x}}\,dx} = \int_{}^{} {\frac{{\cos x - \sin x}}{{{{(\sin x + \cos x)}^2}}}\,dx} $
Now put $\sin x + \cos x = t,$ then the $ \Rightarrow (\cos x - \sin x)\,dx = dt$
required integral is $ - \frac{1}{{\sin x + \cos x}} + c$.
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