Integral Calculus Question 358
Question: $ \int{\cos ({\log_{e}}x),dx} $ is equal to
[MP PET 2003]
Options:
A) $ \frac{1}{2}x{\cos ({\log_{e}}x)+\sin ({\log_{e}}x)} $
B) $ x{\cos ({\log_{e}}x)+\sin ({\log_{e}}x)} $
C) $ \frac{1}{2}x{\cos ({\log_{e}}x)-\sin ({\log_{e}}x)} $
D) $ x{\cos ({\log_{e}}x)-\sin ({\log_{e}}x)} $
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Answer:
Correct Answer: A
Solution:
Let $ I=\int{\cos ({\log_{e}}x),dx} $ $ =\int{\cos ({\log_{e}}x),.,1,dx} $ $ I=\cos ({\log_{e}}x).,x-\int{\frac{-\sin ({\log_{e}}x)}{x}}.,xdx $ $ =x\cos ({\log_{e}}x)+\int{\sin ({\log_{e}}x)}dx $ $ =x\cos ,({\log_{e}}x)+\int{\sin ,({\log_{e}}x)}1dx $ $ =x\cos ({\log_{e}}x)+\sin ({\log_{e}}x).,x-\int{\frac{\cos ({\log_{e}}x)}{x}x,dx} $ $ =x,\cos ({\log_{e}}x)+x\sin ({\log_{e}}x)-I $
$ \Rightarrow 2I=x,[\cos ,({\log_{e}}x)+\sin ,({\log_{e}}x)] $
$ \Rightarrow I=\frac{x}{2},[\cos ({\log_{e}}x)+\sin ,({\log_{e}}x)] $ .
BETA
