∫sin(x2).dx
Substitute u=π2.x
dx=2π.du
∫sin(x2).dx=2π∫sin(2π.u2).du
These is special integral Fresnel integral =S(u)
Plug in solved integrals
2π∫sin(2π.u2).du=2π.S(u)
Undo Substitution u=π2.x
=2π.S(π2.x)+c
Hence
∫sin(x2).dx=2π.S(π2.x)+c
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