Let f(x)=sinx+∑r=02 cosx+2 rπ3
g(x)=cosx+∑r=02 sinx+2 rπ3
h(x)=max{f(x),g(x)}
∫−ππ g{f(x)} sin 2x dx is equalent to
2∫0π g {f(x)} sin 2x dx
4∫0π/2 g {f(x)} sin 2x dx
∫−π/2π/2 cos2x⋅sin5xdx
1
f(x)=sinx+∑r=02 cosx+2rπ3=sinx+cosx+cosx+2π3+cosx+4π3
=sinx+cosx−2⋅12cosx=sinx
similarly g(x) = cosx
∫−ππ g {f(x)} sin2x dx=∫−ππ cos(sinx)⋅sin2x dx=0
since integrand is odd
∫π/4π [ h(x) ] {f(x)}2dx where [.] denotes greatest integer function is
π
3π4
0
5π4
∫π/4π [h(x)]{f(x)}2dx=∫x/4π [sinx]⋅sin2xdx=0
∫0π {g(x)}337dx=
337π2
∫0π {g(x)}337dx=∫0π cos337xdx=∫0π cos337(π−x)dx=−∫0π cos337xdx ∴∫0π cos337xdx=0