Integral Calculus Question 388
Question: If $ x\in ( \frac{\pi }{4},\frac{3\pi }{4} ) $ , then $ \int_{{}}^{{}}{\frac{\sin x-\cos x}{\sqrt{1-\sin 2x}}{e^{\sin x}}\cos x\ dx=} $
Options:
A) $ {e^{\sin x}}+c $
B) $ {e^{\sin x-\cos x}}+c $
C) $ {e^{\sin x+\cos x}}+c $
D) $ {e^{\cos x-\sin x}}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{\sin x-\cos x}{\sqrt{1-\sin 2x}}{e^{\sin x}}\cos x,dx}=\int_{{}}^{{}}{\frac{\sin x-\cos x}{\sin x-\cos x}{e^{\sin x}}\cos x,dx} $ $ =\int_{{}}^{{}}{{e^{\sin x}}\cos x,dx}={e^{\sin x}}+c $ .
BETA
