# integrate 0 to pi [x/(2)^(1/2)+sinx]dx

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

Dear student,

Expand sinx as 1-sin2x/2+sin3x/3+...

And then integrate the result..

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Sagar Singh

B.Tech, IIT Delhi

509 Points
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