Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 42260 by math khazana by abdo last updated on 21/Aug/18

let f(a) = ∫_(−∞) ^(+∞)  cos(ax^2 )dx with a>0  1) calculate f(a) interms of a  ) calculate ∫_(−∞) ^(+∞)    cos(2x^2 )dx  3) find the value of  ∫_(−∞) ^(+∞)  cos(x^2  +x+1)dx .

letf(a)=+cos(ax2)dxwitha>0 1)calculatef(a)intermsofa )calculate+cos(2x2)dx 3)findthevalueof+cos(x2+x+1)dx.

Commented bymaxmathsup by imad last updated on 21/Aug/18

3) ∫_(−∞) ^(+∞)  cos(x^2  +x+1)dx =∫_(−∞) ^(+∞)  cos((x+(1/2))^2  +(3/4))dx  =_(x+(1/2)=((√3)/2)t)       ∫_(−∞) ^(+∞)   cos((3/4)(t^2  +1))((√3)/2) dt  =((√3)/2) ∫_(−∞) ^(+∞)   cos((3/4)t^2  +(3/4))dt  =((√3)/2)  ∫_(−∞) ^(+∞)   {cos((3/4)t^2 )cos((3/4))−sin((3/4)t^2 )sin((3/4))}dt  =((√3)/2) cos((3/4)) ∫_(−∞) ^(+∞)  cos((3/4)t^2 )dt −((√3)/2) sin((3/4))∫_(−∞) ^(+∞)  sin((3/4)t^2 )dt  =((√3)/2) cos((3/4))((√(2π))/(2(√(3/4))))  −((√3)/2) sin((3/4))((√(2π))/(2(√(3/4))))  =((√(2π))/2) cos((3/4))  −((√(2π))/2) sin((3/4)) ⇒  ∫_(−∞) ^(+∞)  cos(x^2  +x +1)dx =(√(π/2)){cos((3/4))−sin((3/4))} .

3)+cos(x2+x+1)dx=+cos((x+12)2+34)dx =x+12=32t+cos(34(t2+1))32dt =32+cos(34t2+34)dt =32+{cos(34t2)cos(34)sin(34t2)sin(34)}dt =32cos(34)+cos(34t2)dt32sin(34)+sin(34t2)dt =32cos(34)2π23432sin(34)2π234 =2π2cos(34)2π2sin(34) +cos(x2+x+1)dx=π2{cos(34)sin(34)}.

Commented bymaxmathsup by imad last updated on 21/Aug/18

we have ∫_(−∞) ^(+∞)  cos(ax^2 )dx−i ∫_(−∞) ^(+∞)  sin(ax^2 )dx  = ∫_(−∞) ^(+∞)   e^(−iax^2 ) dx  =∫_(−∞) ^(+∞)   e^(−((√(ia))x)^2 ) dx =_((√(ia))x =t)     ∫_(−∞) ^(+∞)   e^(−t^2 )  (dt/(√(ia)))  = (1/((√a)e^((iπ)/4) )) (√π)=((√π)/(√a)) e^(−((iπ)/4))   =((√π)/(√a)) {cos((π/4))−isin((π/4))} ⇒  f(a) =((√π)/(√a)) ((√2)/2)  =((√(2π))/(2(√a)))  . also we have ∫_(−∞) ^(+∞)   sin(ax^2 )dx =((√(2π))/(2(√a)))  2) ∫_(−∞) ^(+∞)   cos(2x^2 )dx =f(2) = ((√(2π))/(2(√2)))  =((√π)/2)  .

wehave+cos(ax2)dxi+sin(ax2)dx =+eiax2dx=+e(iax)2dx=iax=t+et2dtia =1aeiπ4π=πaeiπ4=πa{cos(π4)isin(π4)} f(a)=πa22=2π2a.alsowehave+sin(ax2)dx=2π2a 2)+cos(2x2)dx=f(2)=2π22=π2.

Terms of Service

Privacy Policy

Contact: info@tinkutara.com