Integral Calculus Question 486
Question: The value of $ \int_{{}}^{{}}{\frac{\sin x-\cos x}{\sin x+\cos x},dx} $ is
[Pb. CET 2000]
Options:
A) $ \frac{1}{\sin x+\cos x}+c $
B) $ \frac{1}{\sin x-\cos x}+c $
C) $ \log (\sin x+\cos x)+c $
D) $ \log ( \frac{1}{\sin x+\cos x} )+c $
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Answer:
Correct Answer: D
Solution:
Put $ \sin x+\cos x=t\Rightarrow (\cos x-\sin x)dx=dt $
$ \Rightarrow $ $ -(\sin x-\cos x)dx=dt $
$ \therefore $ $ \int_{{}}^{{}}{\frac{\sin x-\cos x}{\sin x+\cos x}},dx=-\int_{{}}^{{}}{\frac{dt}{t}=-\log t+c=\log ( \frac{1}{t} )+c} $ Hence, $ \int_{{}}^{{}}{\frac{\sin x-\cos x}{\sin x+\cos x}dx=\log ( \frac{1}{\sin x+\cos x} )+c} $ .
BETA
