A) \[\frac{x}{2}+\frac{x\cos \,(2\,ln\,x)+2x\,sin(2\operatorname{lnx})}{10}+C\]
B) \[\frac{x}{10}+[5+\cos (2\,\ln \,x)-2\sin (2\,in\,x)]+C\]
C) \[\frac{x}{10}+[5-\cos (2\,\ln \,x)+2\sin (2\,in\,x)]+C\]
D) None of these
Correct Answer: A
Solution :
\[f(x)=I=\int{\left\{ \frac{1+\cos \,(2\,In\,x)}{2} \right\}}dx\] \[=\frac{x}{2}+\frac{1}{2}\int{\cos (2\,In\,x)dx=\frac{x}{2}}+\frac{1}{2}{{I}_{1}}\] \[{{I}_{1}}=\int{\cos \,(2\,\,In\,\,x)\,dx=\int{\cos \,\,(2\,\,In\,x).1dx}}\] Integrating by parts using 1 as second function \[=x\cos (2\,\,In\,\,x)+\int{x\sin (2\,\,In\,\,x})\frac{2}{x}\,dx\] \[=x\cos (2\,\,In\,\,x)+2\int{\sin }(2\,\,In\,\,x)dx\] \[=x\cos (2\,\,In\,\,x)+2\{x\sin (2\,\,In\,\,x)-4\int{\cos }(2\,\,In\,\,x)dx\]\[\Rightarrow \,\,5{{I}_{1}}=x\cos (2\,\,In\,\,x)+2x\sin (2\,\,In\,\,x)\] \[\Rightarrow \,\,{{I}_{1}}=\frac{x}{5}[\cos \,\,(2\,\,In\,\,x)+2sin(2\,\,In\,\,x)]\] \[I=f(x)=\frac{x}{2}+\frac{x\cos (2\,\,In\,\,x)+2x\sin (2\,\,In\,\,x)}{10}+C\]
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