yungman Messages 5,741 Reaction score 291 Thread starter Sep 8, 2011 #1 Can anyone show me how to do this: [tex]\int_a^b \left[\frac {\sin x}{x}\right]^2 dx[/tex]
alexfloo Messages 192 Reaction score 0 Sep 8, 2011 #2 It depends on you purpose, but in general the best bet may be power series methods. The Taylor series for sin2 is [itex]\sin^2 x = -\sum^\infty_{k=1} \frac{(-1)^{k} 2^{2k-1}x^{2k}}{(2k)!}[/itex] So we have [itex]\frac{\sin^2 x}{x^2} = -\sum^\infty_{k=1} \frac{(-1)^{k} 2^{2k-1}x^{2k-2}}{(2k)!}[/itex] Then you can try term-wise integration and the fundamental theorem of calculus to get a general form for the answer as a power series.
It depends on you purpose, but in general the best bet may be power series methods. The Taylor series for sin2 is [itex]\sin^2 x = -\sum^\infty_{k=1} \frac{(-1)^{k} 2^{2k-1}x^{2k}}{(2k)!}[/itex] So we have [itex]\frac{\sin^2 x}{x^2} = -\sum^\infty_{k=1} \frac{(-1)^{k} 2^{2k-1}x^{2k-2}}{(2k)!}[/itex] Then you can try term-wise integration and the fundamental theorem of calculus to get a general form for the answer as a power series.