intrigate 1/sinx(3+2cosx)

1 ANSWERS

1. paafamily
   

   we know that d/dx cos x = - sin x so
   muliply both numerator and denominator by sin x
   
   to get sin x/(1-cos^2 x) (3+ 2 cos x) as sin ^2 x = 1- cos^2 x
   
   put cos x= t to get integral
   
   = - dt/(1-t^2)(3+2t) = - dt/(1-t)(1+t)(3+2t)
   
   convert to partial fraction
   
   1/(1-t)(1+t)(3+2t) = A/(1-t) + B/(1+t) + C/(3+2t)
   
   or 1 = A(1+t)(3+2t) + B(1-t)(3+2t) + C(1+t)(1-t)
   
   t = - 1 gives 1 = -2 B or B = - 1/2
   t = 1 gives 1 = 10 A or A = 1/10
   T = 0 gives 1 = 3A + 3 B + C or C = 1 - 3/10 + 3/2 = 22/10
   
   so 1/(1-t)(1+t)(3+2t) = 1/(10(1-t)) - 1/(2(1+t)) + 11/5((3+2t))
   
   or given expression = - (1/(10(1-t)) - 1/(2(1+t)) + 11/5((3+2t))
   
   intgrating we get 1/10 ln |1-t| - 1/2 ln |1+t| + 11/10 ln | 3/2+t| where t = cos x

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