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Question Number 75830 by Master last updated on 18/Dec/19

Commented by Master last updated on 18/Dec/19

sir mind is power , sir mathmax plz

$$\mathrm{sir}\:\mathrm{mind}\:\mathrm{is}\:\mathrm{power}\:,\:\mathrm{sir}\:\mathrm{mathmax}\:\mathrm{plz} \\ $$

Commented by turbo msup by abdo last updated on 18/Dec/19

do you mean (sinx)^2  or sin(x^2 )?

$${do}\:{you}\:{mean}\:\left({sinx}\right)^{\mathrm{2}} \:{or}\:{sin}\left({x}^{\mathrm{2}} \right)? \\ $$

Commented by Master last updated on 18/Dec/19

sin(x^2 )

$$\mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} \right) \\ $$

Commented by mathmax by abdo last updated on 19/Dec/19

at form of serie  we have  sin(x^2 ) =Σ_(n=0) ^∞  (((−1)^n (x^2 )^(2n+1) )/((2n+1)!))  =Σ_(n=0) ^∞  (((−1)^n x^(4n+2) )/((2n+1)!)) ⇒ ((sin(x^2 ))/x^2 ) =Σ_(n=0) ^∞  (((−1)^n )/((2n+1)!))x^(4n)  ⇒  ∫  ((sin(x^2 ))/x^2 )dx =Σ_(n=0) ^∞  (((−1)^n )/((2n+1)(4n+1)))x^(4n+1)  +C

$${at}\:{form}\:{of}\:{serie}\:\:{we}\:{have}\:\:{sin}\left({x}^{\mathrm{2}} \right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \left({x}^{\mathrm{2}} \right)^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{4}{n}+\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\Rightarrow\:\frac{{sin}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{x}^{\mathrm{4}{n}} \:\Rightarrow \\ $$$$\int\:\:\frac{{sin}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{4}{n}+\mathrm{1}\right)}{x}^{\mathrm{4}{n}+\mathrm{1}} \:+{C}\: \\ $$

Commented by abdomathmax last updated on 19/Dec/19

let f(x)=∫_0 ^x  ((sin(t^2 ))/t^2 )dt ⇒  f(x)=[−((sin(t^2 ))/t)]_0 ^x −∫_0 ^x (−(1/t))(2t cos(t^2 )dt  =−((sin(x^2 ))/x) +2 ∫_0 ^x  cos(t^2 )dt  ∫_0 ^x  cos(t^2 )dt =Re(∫_0 ^x  e^(−it^2 ) dt)  and  changement  it^2 =u give t^2 =−iu ⇒t =(−iu)^(1/2)  ⇒  dt =(−i)^(1/2) (1/2)u^(−(1/2))  =(1/2)e^(−((iπ)/4))   u^(−(1/2))  ⇒  ∫_0 ^x   e^(−it^2 ) dt =(1/2)e^(−((iπ)/4)) ∫_0 ^(ix^2 )   e^(−u)   u^(−(1/2))  du  ...be continued...

$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }{dt}\:\Rightarrow \\ $$$${f}\left({x}\right)=\left[−\frac{{sin}\left({t}^{\mathrm{2}} \right)}{{t}}\right]_{\mathrm{0}} ^{{x}} −\int_{\mathrm{0}} ^{{x}} \left(−\frac{\mathrm{1}}{{t}}\right)\left(\mathrm{2}{t}\:{cos}\left({t}^{\mathrm{2}} \right){dt}\right. \\ $$$$=−\frac{{sin}\left({x}^{\mathrm{2}} \right)}{{x}}\:+\mathrm{2}\:\int_{\mathrm{0}} ^{{x}} \:{cos}\left({t}^{\mathrm{2}} \right){dt} \\ $$$$\int_{\mathrm{0}} ^{{x}} \:{cos}\left({t}^{\mathrm{2}} \right){dt}\:={Re}\left(\int_{\mathrm{0}} ^{{x}} \:{e}^{−{it}^{\mathrm{2}} } {dt}\right)\:\:{and}\:\:{changement} \\ $$$${it}^{\mathrm{2}} ={u}\:{give}\:{t}^{\mathrm{2}} =−{iu}\:\Rightarrow{t}\:=\left(−{iu}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\Rightarrow \\ $$$${dt}\:=\left(−{i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}}{u}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:{u}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{{x}} \:\:{e}^{−{it}^{\mathrm{2}} } {dt}\:=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \int_{\mathrm{0}} ^{{ix}^{\mathrm{2}} } \:\:{e}^{−{u}} \:\:{u}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{du} \\ $$$$...{be}\:{continued}... \\ $$

Answered by mr W last updated on 18/Dec/19

=∫sin x^2  d(−(1/x))  =−((sin x^2 )/x)+2∫cos x^2 dx  =−((sin x^2 )/x)+(√(2π))∫cos [(π/2)((((√2)x)/(√π)))^2 ]d((((√2)x)/(√π)))  =−((sin x^2 )/x)+(√(2π)) C((((√2)x)/(√π)))+C  with Fresnel integral:  S(x)=∫_0 ^x sin (((πt^2 )/2))dt  C(x)=∫_0 ^x cos (((πt^2 )/2))dt

$$=\int\mathrm{sin}\:{x}^{\mathrm{2}} \:{d}\left(−\frac{\mathrm{1}}{{x}}\right) \\ $$$$=−\frac{\mathrm{sin}\:{x}^{\mathrm{2}} }{{x}}+\mathrm{2}\int\mathrm{cos}\:{x}^{\mathrm{2}} {dx} \\ $$$$=−\frac{\mathrm{sin}\:{x}^{\mathrm{2}} }{{x}}+\sqrt{\mathrm{2}\pi}\int\mathrm{cos}\:\left[\frac{\pi}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}{x}}{\sqrt{\pi}}\right)^{\mathrm{2}} \right]{d}\left(\frac{\sqrt{\mathrm{2}}{x}}{\sqrt{\pi}}\right) \\ $$$$=−\frac{\mathrm{sin}\:{x}^{\mathrm{2}} }{{x}}+\sqrt{\mathrm{2}\pi}\:{C}\left(\frac{\sqrt{\mathrm{2}}{x}}{\sqrt{\pi}}\right)+{C} \\ $$$${with}\:{Fresnel}\:{integral}: \\ $$$${S}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \mathrm{sin}\:\left(\frac{\pi{t}^{\mathrm{2}} }{\mathrm{2}}\right){dt} \\ $$$${C}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \mathrm{cos}\:\left(\frac{\pi{t}^{\mathrm{2}} }{\mathrm{2}}\right){dt} \\ $$

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