theaviator Messages 6 Reaction score 0 Thread starter Sep 21, 2007 #1 Y=4 x=2 ∫ ∫ Sinx ∕ x^2 dx dy Y=0 x=√y
Astronuc Staff Emeritus Science Advisor Gold Member 2025 Award Messages 22,714 Reaction score 7,805 Sep 21, 2007 #2 Can one find an integral table with [tex]\int \frac{sin\,x}{x^2}\,dx[/tex] ? Alternatively, expand sin x as an infinite series, divide by x2 and solve for each term.
Can one find an integral table with [tex]\int \frac{sin\,x}{x^2}\,dx[/tex] ? Alternatively, expand sin x as an infinite series, divide by x2 and solve for each term.
theaviator Messages 6 Reaction score 0 Sep 22, 2007 #3 i changed the order of the integration...the final answer i had is the same one in the sheet from which i get the problem...but i don't know id the way i solved it is right or wrong...please HELP... X=2 y=x^2 ∫ ∫ sinx/x^2 dx dy X=0 y=0 X=2 y=x^2 = ∫ [(sinx/x^2)y] dx X=0 y=0 X=2 =∫ sinx dx X=0 X=2 = [ -cosx] X=0 = -cos2+cos(0)= 1-cos2
i changed the order of the integration...the final answer i had is the same one in the sheet from which i get the problem...but i don't know id the way i solved it is right or wrong...please HELP... X=2 y=x^2 ∫ ∫ sinx/x^2 dx dy X=0 y=0 X=2 y=x^2 = ∫ [(sinx/x^2)y] dx X=0 y=0 X=2 =∫ sinx dx X=0 X=2 = [ -cosx] X=0 = -cos2+cos(0)= 1-cos2
nicksauce Science Advisor Homework Helper Messages 1,270 Reaction score 8 Sep 22, 2007 #4 That is correct.