Integral Calculus Question 6
Question: $ \int_{{}}^{{}}{\frac{\sin 2xdx}{1+{{\cos }^{2}}x}}= $
[Karnataka CET 2005]
Options:
A) $ \frac{1}{2}\log (1+{{\cos }^{2}}x)+c $
B) $ 2\log (1+{{\cos }^{2}}x)+c $
C) $ \frac{1}{2}\log (1+\cos 2x)+c $
D) $ -\log (1+{{\cos }^{2}}x)+c $
Show Answer
Answer:
Correct Answer: D
Solution:
$ I=\int{\frac{\sin 2x}{1+{{\cos }^{2}}x}dx=\int{\frac{2\sin x\cos x}{1+{{\cos }^{2}}x}dx}} $ Put $ 1+{{\cos }^{2}}x=t $
Þ $ -2\sin x\cos xdx=dt $
Þ $ \sin 2x=-dt $ . Hence $ I=\int{^{-}( \frac{dt}{t} )}=-\log t+c $ $ =-\log (1+{{\cos }^{2}}x)+c $ .
BETA
