Integral Calculus Question 314
Question: $ \int_{{}}^{{}}{\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }}dx= $
[MP PET 1994]
Options:
A) $ 2[\sin x+x\cos \alpha ]+c $
B) $ 2[\sin x+\sin \alpha ]+c $
C) $ 2[-\sin x+x\cos \alpha ]+c $
D) $ -2[\sin x+\sin \alpha ]+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }},dx=\int_{{}}^{{}}{\frac{2({{\cos }^{2}}x-{{\cos }^{2}}\alpha )}{\cos x-\cos \alpha },dx} $ $ =2\int_{{}}^{{}}{(\cos x+\cos \alpha ),dx}=2(\sin x+x\cos \alpha ) $ .
BETA
