A) \[\frac{1}{2}x[\cos (\log x)-\sin (\log x)]\]
B) \[\cos (\log x)-x\]
C) \[\frac{1}{2}x[\sin (\log x)-\cos (\log x)]\]
D) \[-\cos \log x\]
Correct Answer: C
Solution :
Let \[I=\int_{{}}^{{}}{\sin (\log x)\,dx}\] Put \[\log x=t\Rightarrow x={{e}^{t}}\Rightarrow dx={{e}^{t}}dt,\] then \[I=\int_{{}}^{{}}{\sin t\,.\,{{e}^{t}}\,dt}=\sin t\,.\,{{e}^{t}}-\int_{{}}^{{}}{{{e}^{t}}.\,\cos t\,dt}\] \[=\sin t\,.\,{{e}^{t}}-[\cos t\,.\,{{e}^{t}}+\int_{{}}^{{}}{{{e}^{t}}.\sin t\,dt}]\] \[\Rightarrow 2I=\sin t\,.\,{{e}^{t}}-\cos t\,.\,{{e}^{t}}\] \[\Rightarrow I=\int_{{}}^{{}}{\sin \,(\log x)\,dx}\]\[=\frac{1}{2}x\,[\sin \,(\log x)-\cos \,(\log x)]\].
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