Integral Calculus Question 405
Question: $ \int_{{}}^{{}}{\frac{\cos 2x}{{{(\cos x+\sin x)}^{2}}}\ dx=} $
Options:
A) $ \log \sqrt{\cos x+\sin x}+c $
B) $ \log (\cos x-\sin x)+c $
C) $ \log (\cos x+\sin x)+c $
D) $ -\frac{1}{\cos x+\sin x}+c $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \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}}}}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}},dt=\log t+c=\log (\sin x+\cos x)+c $ .
BETA
