A) \[0\]
B) \[x+c\]
C) \[\frac{x}{2}-\frac{\cos \pi x}{2\pi }+c\]
D) \[\frac{1}{1+{{\pi }^{x}}}\frac{{{\cos }^{2}}\pi x}{2\pi }+c\]
E) \[\frac{x}{2}-\frac{\sin 2\pi x}{4\pi }+c\]
Correct Answer: E
Solution :
\[f(x)=\frac{{{\sin }^{2}}\pi x}{1+{{\pi }^{x}}}\] Now, \[f(x)+f(-x)=\frac{{{\sin }^{2}}\pi x}{1+{{\pi }^{x}}}+\frac{{{\sin }^{2}}(-\pi x)}{1+{{\pi }^{-x}}}\] \[=\frac{{{\sin }^{2}}\pi x}{1+{{\pi }^{x}}}+\frac{({{\sin }^{2}}\pi x){{\pi }^{x}}}{{{\pi }^{x}}+1}\] \[=\frac{{{\sin }^{2}}\pi x}{1+{{\pi }^{x}}}(1+{{\pi }^{x}})={{\sin }^{2}}\pi x\] \[=\frac{1-\cos 2\pi x}{2}\] \[\therefore \]\[\int{(f(x)+f(-x))}\,dx=\int{\left[ \frac{1-\cos 2\pi x}{2} \right]}\,dx\] \[=\frac{1}{2}x-\frac{\sin 2\pi x}{4\pi }+x\]
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