If f(x)=sin2πx1+πx. Then, ∫f(x)+f(-x)dx is equal to
0
x+C
x2-cosπx2π+C
11+πcos2πx2π+C
x2-sin2πx4π+C
Explanation for the correct option:
Find the value of ∫f(x)+f(-x)dx:
Given,
f(x)=sin2πx1+πx...1
f(-x)=sin2-πx1+π-x=πxsin2-πx1+πx...2
Substitute equations (1) and (2) in ∫f(x)+f(-x)dx.
∫f(x)+f(-x)dx=∫sin2πx1+πx+πxsin2πx1+πxdx=∫sin2πx1+πx1+πxdx=∫sin2πxdx=12∫1-cos2πxdx[∵2sin2πx=1-cos2πx]=12x-sin2πx2π+C=x2-sin2πx4π+C
Hence, the correct option is E.