vikas askiitian expert
Last Activity: 13 Years ago
I = x/(21/2+sinx) ..........1 lim 0 to pi
{ f(x)dx from a to b = f(a+b-x) lim from a to b } (property )
by using this property I becomes
I ={ (pi-x)/21/2+sin(pi-x) }dx = pi-x/(21/2+sinx) lim 0 to pi ............2 (sinpi-x = sinx)
adding 1 & 2
2I = {pi/21/2+sinx }dx lim 0 to pi
now , sinx = 2tan(x/2)/1+tan2(x/2) (formula)
putting this in above
2I = pi{1+tan2x/2)/(21/2+21/2tan2x/2 + 2tanx/2) dx lim 0 to pi
1+tan2x/2 = sec2x/2 (formula)
2I = pi{sec2x/2/(21/2 +21/2tan2x/2 + 2tanx/2) } dx lim 0 to pi
now put tanx/2 = t then , sec2x/2dx = 2dt
2I ={ 2pi/(21/2+21/2t2+2t) } dt lim 0 to infinity
I = pi/21/2 . {1/t2+21/2t+1} dt lim 0 to infinity
now this becomes simple integral and u can easily proceed now...
approve if u like it