Integral Calculus Question 58
Question: $ \int{\frac{dx}{7+5\cos x}=} $
[EAMCET 2002]
Options:
A) $ \frac{1}{\sqrt{6}}{{\tan }^{-1}}( \frac{1}{\sqrt{6}}\tan \frac{x}{2} )+c $
B) $ \frac{1}{\sqrt{3}}{{\tan }^{-1}}( \frac{1}{\sqrt{3}}\tan \frac{x}{2} )+c $
C) $ \frac{1}{4}{{\tan }^{-1}}( \tan \frac{x}{2} )+c $
D) $ \frac{1}{7}{{\tan }^{-1}}( \tan \frac{x}{2} )+c $
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Answer:
Correct Answer: A
Solution:
$ I=\frac{dx}{7+5\cos x} $ $ =\int{\frac{dx}{7+5,( \frac{1-{{\tan }^{2}}(x/2)}{1+{{\tan }^{2}}(x/2)} )}} $ $ =\int{\frac{{{\sec }^{2}}(x/2),dx}{7+7{{\tan }^{2}}(x/2)+5-5{{\tan }^{2}}(x/2)}} $ $ =\int{\frac{{{\sec }^{2}}(x/2),dx}{12+2{{\tan }^{2}}(x/2)}} $ $ =\int{\frac{\frac{1}{2}{{\sec }^{2}}(x/2),.dx}{6+{{\tan }^{2}}(x/2)}} $ Put $ \tan \frac{x}{2}=t $
Þ $ \frac{1}{2}{{\sec }^{2}}\frac{x}{2}dx=dt $ $ I=\int{\frac{dt}{t^{2}+({{\sqrt{6)}}^{2}}}} $ $ =\frac{1}{\sqrt{6}}{{\tan }^{-1}}\frac{t}{\sqrt{6}}+c $ $ =\frac{1}{\sqrt{6}}{{\tan }^{-1}}| \frac{\tan (x/2)}{\sqrt{6}} |+c $ .
BETA
