MCQ
$\int_{}^{} {\frac{{\cos 2x}}{{{{(\cos x + \sin x)}^2}}}\;dx = } $
- A$\log \sqrt {\cos x + \sin x} + c$
- B$\log (\cos x - \sin x) + c$
- ✓$\log (\cos x + \sin x) + c$
- D$ - \frac{1}{{\cos x + \sin x}} + c$
✓
Answer
Correct option: C.
$\log (\cos x + \sin x) + c$
c
(c)$\int_{}^{} {\frac{{\cos 2x}}{{{{(\cos x + \sin x)}^2}}}dx} = \int_{}^{} {\frac{{(\cos x - \sin x)(\cos x + \sin x)}}{{{{(\cos x + \sin x)}^2}}}{\rm{ }}} dx$
$ = \int_{}^{} {\frac{{\cos x - \sin x}}{{\cos x + \sin x}}dx} $
Put $t = \sin x + \cos x \Rightarrow dt = (\cos x - \sin x)dx$,
(c)$\int_{}^{} {\frac{{\cos 2x}}{{{{(\cos x + \sin x)}^2}}}dx} = \int_{}^{} {\frac{{(\cos x - \sin x)(\cos x + \sin x)}}{{{{(\cos x + \sin x)}^2}}}{\rm{ }}} dx$
$ = \int_{}^{} {\frac{{\cos x - \sin x}}{{\cos x + \sin x}}dx} $
Put $t = \sin x + \cos x \Rightarrow dt = (\cos x - \sin x)dx$,
then it reduces to $\int_{}^{} {\frac{1}{t}} \,{\rm{ }}dt = \log t + c = \log (\sin x + \cos x) + c$.
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