Integral Calculus Question 276
Question: If $ \int_{{}}^{{}}{(\cos x-\sin x)\ dx=\sqrt{2}\sin (x+\alpha )+c} $ , then $ \alpha = $
Options:
A) $ \frac{\pi }{3} $
B) $ -\frac{\pi }{3} $
C) $ \frac{\pi }{4} $
D) $ -\frac{\pi }{4} $
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Answer:
Correct Answer: C
Solution:
Given that $ \int_{{}}^{{}}{(\cos x-\sin x),dx}=\sqrt{2}\sin (x+\alpha )+c $
$ \Rightarrow \sin x+\cos x+c=\sqrt{2}\sin (x+\alpha )+c $
$ \Rightarrow \sqrt{2}( \frac{\sin x}{\sqrt{2}}+\frac{\cos x}{\sqrt{2}} )+c=\sqrt{2}\sin (x+\alpha )+c $
$ \Rightarrow \sqrt{2}\sin ( x+\frac{\pi }{4} )+c=\sqrt{2}\sin (x+\alpha )+c $
$ \Rightarrow \sin ( x+\frac{\pi }{4} )=\sin (x+\alpha )\Rightarrow \alpha =\frac{\pi }{4}. $
BETA
