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vikash chandoa Grade: 12
        

what is integration of 1/(a^2-2*a*cos(x)+1) its lower limit is 0 and upper limit is pie

6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

I= dx/[a2 -2acosx +1]       lim 0 to pi


put cosx = 1-tan2x/2/1+tan2x/2


I= sec2x/2 dx / [(a-1)2 + (a+1)2tan2x/2]


now put tanx/2 = t


    sec2x/2dx =2dt


I = 2dt / [(a-1)2+(a+1)2t2)           


I =2dt/(1+a)2 .[(a-1/a+1)2 + t2]


this integral is same as  1/a2+x2 dx & its integral is tan-1(x/a)/a


so


 I =2/(a2-1)tan-1[(a+1)t/(a-1)]     


 I = 2/(a2-1)tan-1[(a+1)tanx/2/(a-1)]               lim 0 tp pi


taking limit


 I =2/(a2-1) [ tan-1infinity -tan-10]


    =pi/(a2-1)         ans

6 years ago
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