Integral Calculus Question 123
Question: $ \int{\frac{dx}{\cos x+\sqrt{3}\sin x}} $ equals
Options:
A) $ \log \tan ( \frac{x}{2}+\frac{\pi }{12} )+C $
B) $ \log \tan ( \frac{x}{2}-\frac{\pi }{12} )+C $
C) $ \frac{1}{2}\log \tan ( \frac{x}{2}+\frac{\pi }{12} )+C $
D) $ \frac{1}{2}\log \tan ( \frac{x}{2}-\frac{\pi }{12} )+C $
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Answer:
Correct Answer: C
Solution:
[c] $ I=\int{\frac{dx}{\cos x+\sqrt{3}\sin x}} $
$ \Rightarrow I=\int{\frac{dx}{2[ \frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x ]}} $ $ =\frac{1}{2}\int{\frac{dx}{[ \sin \frac{\pi }{6}\cos x+\cos ,\frac{\pi }{6}\sin x ]}} $ $ =\frac{1}{2}.\int{\frac{dx}{\sin ( x+\frac{\pi }{6} )}} $
$ \Rightarrow I=\frac{1}{2}.\int{\cos ec( x+\frac{\pi }{6} )dx} $ $ \because \int{\cos ecxdx=\log | (\tan x/2) |+C} $
$ \therefore I=\frac{1}{2}.\log \tan ( \frac{x}{2}+\frac{\pi }{12} )+C $
BETA
