Integral of sin(x^2)
∫sin(x2).dx\smallint{sin(x^2)}.dx∫sin(x2).dx
Substitute u=2.xπu=\frac{\sqrt{2}.x}{\sqrt{π}}u=π2.x
dx=π2.dudx=\frac{\sqrt{π}}{\sqrt{2}}.dudx=2π.du
∫sin(x2).dx=π2∫sin(π.u22).du\smallint{sin(x^2)}.dx= \frac{\sqrt{π}}{\sqrt{2}}\smallint{sin(\frac{π.u^2}{2})}.du∫sin(x2).dx=2π∫sin(2π.u2).du
These is special integral Fresnel integral =S(u)=S(u)=S(u)
Plug in solved integrals
π2∫sin(π.u22).du=π.S(u)2\frac{\sqrt{π}}{\sqrt{2}}\smallint{sin(\frac{π.u^2}{2})}.du=\frac{\sqrt{π}.S(u)}{\sqrt2}2π∫sin(2π.u2).du=2π.S(u)
Undo Substitution u=2.xπu=\frac{\sqrt{2}.x}{\sqrt{π}}u=π2.x
=π.S(2.xπ)2+c=\frac{\sqrt{π}.S( \frac{\sqrt{2}.x}{\sqrt{π}} )}{\sqrt2}+c=2π.S(π2.x)+c
Hence
∫sin(x2).dx=π.S(2.xπ)2+c\smallint{sin(x^2)}.dx =\frac{\sqrt{π}.S( \frac{\sqrt{2}.x}{\sqrt{π}} )}{\sqrt2}+c∫sin(x2).dx=2π.S(π2.x)+c
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