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definite integration of sin^2 x/sin x+cos x from 0 to pie
integral sin^2x/sinx+cosx from 0 to pie =
integral{( sin^2x)/(sinx+cosx) + (cos^2x)/(sinx+cosx)}from 0to pie/2 after applying propertie of integral
=integral( 1/sinx+cosx)integral from 0 t0 pie/2
= put sinx=2tan(x/2)/(1+tan^x/2) and cosx=(1-tan^x/2)/(1+tan^2x/2)
now expression becomes integral (sex^2x/2) /(1+2tan(x/2) -tan^2x/2) limit from 0 to pie/2
now put tanx/2 = t
we get sec^2dx=2dt
expression becomes integral 2dt/1+2t-t^2 limit 0 to 1
=integral -2/t^2 - 2 limit 0 to 1
=(1/underoot2) .(log(underoot2- 1)/(underoot2 +1))
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