Here's the question you clicked on:
shahzadjalbani
integrate sin (x^2) form 0 to pi
\[\int\limits_{0}^{\pi} \sin x ^{2} dx\]
You cannot do that in closed from.
You can do it numerically.
first substitute x^2=z then 2xdx=dz putting x=sqrt(z) it will be dx=dz/(2sqrt(z)) changing limits and integrating by parts should give you the answer
the best option is to go for numerical integration f(x)=sin(x^2) using simpson's rule with n=4 h=(pi-0)/4=pi/4 f(0)=sin(0)=0 f(pi/4)=sin(pi/4^2)=0.5784 f(pi/2)=sin(pi/2^2)=0.6242 f(3pi/4)=sin(3pi/4^2)=-0.668 f(pi)=sin(pi^2)=-0.4303 using sinpsons rule h/3(f(0)+4f(pi/4)+2f(pi/2)+4f(3pi/4)+f(pi)) put all the above values =0.12044
@sami-21 Can you please explain to me what is simpson's rule?
Sami-21, just wondering, why does other sources suggest that the answer is 0.77265 instead?
yeah sure it is a technique for numerical integration the basic formula for this rule is as follows \[\int\limits_{a}^{b}f(x)=h/3(f0(x)+4f1(x)+2f2(x)+4f3(x)+2f4(x)+...fn(x))\] where h is the size of the interval and given by h=(b-a)/n where n is the no of subintervals (no of rectangles that gets added) just evaluate the function at the respective points the point with odd subscripts is multiplied by 4 and that with even subscripts is multiplied by 2 you can see this in the above formula you can visit the following also for detail http://en.wikipedia.org/wiki/Simpson's_rule
@sami-21 Big thanks man!!
@cherylim23 yes correct answer is 0.77265 .it is a numerical technique there is always eror associated with numerical techniques we go for numerical technique when we do not have analytical solution ..the same answer can be achieved by simpson rule with n=20 or more you can go there at the following site and can compute the result with different size http://nastyaccident.com/calculators/calculus/simpsonsRule
here is a very nice integral and possible to solve analytically ; if u r interested \[\large \int\limits_{0}^{\infty} \sin x^2 \ dx\]
yes i can do this !!!!!!
taylor series about a=0 for sin|(x) is \[sin(x)=x-x^3/3!+x^5/5!-x^7/7!+...\] replace x by x^2 i the above \[sin(x^2)=x^2-x^6/3!+x^{10}/5!-x^{14}/7!+...\] now you can integrate both sides and you will get the result because you will have just to integrate polynomial
yes thats right it will give a series for u but not the exact answer here is my solution ; It worths watching! :)
yup i got this in BS grewall Highier Engineering Mathematics thanks i think fourier integral also works here
yes fourier will work also....