## How to integrate Sin(x)/(x)?

you need to use the midpoint rule, the trapezium rule and then apply these to simpsons rule silly, what's the width ur integrating?

 open your MATLAB and insert the operation then differentiate the answer given by MATLAB then search the solution reverse.... (start from the final step and come to the first step) sometimes the above method will work...but even I am not sure about it
 There could be other way as to go for fourier transform keeping f=0 (frequency )in its equation.If h(x)=\hat{f}(x) then  \hat{h}(\xi)= f(-\xi). that is duality of fourier transform.
 well.... i looked through all the replies to this question and felt that no one really answered it. sooooo i got the integral to sin(x)/(x) to = -cos(x)(1/2x) + (1/2)ln(x)sin(x) +c done by integration by parts twice. so integral of udv = uv - integral of vdu u = lnx du = 1/x dv = sinx v = -cosx u get -cosx/x + (integral of cosxlnx) ------ do integration by parts again u = lnx du = 1/x dv = cosx v = sinx so u get lnxsinx - (integral of sinx/x). add the (integral of sinx/x) over. So now you have 2(integral of sinx/x) = -cosx/x + lnxsinx divide the 2 over n you get -cos(x)(1/2x) + (1/2)ln(x)sin(x) +c

 Quote by Mr.Rabbit87 well.... i looked through all the replies to this question and felt that no one really answered it. sooooo i got the integral to sin(x)/(x) to = -cos(x)(1/2x) + (1/2)ln(x)sin(x) +c done by integration by parts twice. so integral of udv = uv - integral of vdu u = lnx du = 1/x dv = sinx v = -cosx u get -cosx/x + (integral of cosxlnx) ------ do integration by parts again u = lnx du = 1/x dv = cosx v = sinx so u get lnxsinx - (integral of sinx/x). add the (integral of sinx/x) over. So now you have 2(integral of sinx/x) = -cosx/x + lnxsinx divide the 2 over n you get -cos(x)(1/2x) + (1/2)ln(x)sin(x) +c
i lied... i differentiate wrong lol