Integral Calculus Question 496
Question: $ \int{\frac{\sin xdx}{3+4{{\cos }^{2}}x}=} $
[Karnataka CET 2000]
Options:
A) $ \log (3+4{{\cos }^{2}}x)+c $
B) $ \frac{-1}{2\sqrt{3}}{{\tan }^{-1}}( \frac{\cos x}{\sqrt{3}} )+c $
C) $ \frac{-1}{2\sqrt{3}}{{\tan }^{-1}}( \frac{2\cos x}{\sqrt{3}} )+c $
D) $ \frac{1}{2\sqrt{3}}{{\tan }^{-1}}( \frac{2\cos x}{\sqrt{3}} )+c $
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Answer:
Correct Answer: C
Solution:
$ I=\int{\frac{\sin x}{3+4{{\cos }^{2}}x}dx} $ Put $ \cos x=t $
Þ $ -\sin xdx=dt $
$ \therefore I=\int{\frac{-dt}{3+4t^{2}}} $ $ =\int{\frac{{{\sec }^{2}}x}{1-{{\tan }^{2}}x}dx} $
Þ $ I=-\frac{1}{4.\frac{\sqrt{3}}{2}}.{{\tan }^{-1}}\frac{t}{( \frac{\sqrt{3}}{2} )}+c=\frac{-1}{2\sqrt{3}}{{\tan }^{-1}}\frac{2,t}{\sqrt{3}}+c $
$ \Rightarrow I=\frac{-1}{2\sqrt{3}}{{\tan }^{-1}}( \frac{2\cos x}{\sqrt{3}} )+c $ .
BETA
