A) \[\frac{x}{2}[\sin (\log x)-\cos (\log x)]+c\]
B) \[\frac{x}{2}[\sin (\log x)+\cos (\log x)]+c\]
C) \[\frac{1}{2}[\sin (\log x)-\cos (\log x)]+c\]
D) \[\frac{1}{2}[\sin (\log x)+\cos (\log x)]+c\]
Correct Answer: B
Solution :
Let \[I=\int{\cos (\log x)dx}=\int{\underset{I}{\mathop{\cos }}\,(\log x).\underset{II}{\mathop{1}}\,dx}\] \[\Rightarrow \] \[I=\cos (\log x).x+\int{\frac{\sin (\log x)}{x}}.xdx\] \[\Rightarrow \] \[I=x.\cos (\log x)+\sin (\log x).x\] \[-\int{\frac{\cos (\log x)}{x}.x\,dx}\] \[\Rightarrow \]\[I=x.\cos (\log x)+x\sin (\log x)-I+c\] \[\Rightarrow \]\[2I=x[\cos (\log x)+\sin (\log x)+c\] \[\Rightarrow \]\[I=\frac{x}{2}[\sin (\log x)+\cos (\log x)]+c\]
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