Integral Calculus Question 193
Question: $ \int_{{}}^{{}}{
[\sin (\log x)+\cos (\log x)]}\ dx= $ [MP PET 1991]
Options:
A) $ x\cos (\log x)+c $
B) $ \sin (\log x)+c $
C) $ \cos (\log x)+c $
D) $ x\sin (\log x)+c $
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Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{\sin (\log x),dx}+\int_{{}}^{{}}{\cos (\log x),dx} $ $ =x\sin (\log x)-\int_{{}}^{{}}{\frac{x\cos (\log x)}{x}},dx+\int_{{}}^{{}}{\cos (\log x),dx+c} $ $ =x\sin (\log x)+c. $
BETA
