A) \[\sqrt{2}\]
B) \[2\sqrt{2}\]
C) \[4\sqrt{2}\]
D) \[8\sqrt{2}\]
Correct Answer: D
Solution :
\[I=\int\limits_{-4\pi \sqrt{2}}^{4\pi \sqrt{2}}{\left( \frac{\sin x}{1+{{x}^{4}}}+1 \right)dx}\,=\int\limits_{-4\pi \sqrt{2}}^{4\pi \sqrt{2}}{\frac{\sin x}{1+{{x}^{4}}}dx+\int\limits_{-4\pi \sqrt{2}}^{4\pi \sqrt{2}}{1dx}}\] Since the function \[\frac{\sin x}{1+{{x}^{4}}}\] is odd we get that \[\int\limits_{-4\pi \sqrt{2}}^{4\pi \sqrt{2}}{\left( \frac{\sin x}{1+{{x}^{4}}}\, \right)dx=0}\] Hence, \[I=\int\limits_{-4\pi \sqrt{2}}^{4\pi \sqrt{2}}{1dx\Rightarrow \,I=8\pi \sqrt{2}\,\,\equiv \,k\pi }\] \[\Rightarrow \,k=8\sqrt{2}\]
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