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Chilukuri Sai Kartik Grade: 12
        

∫cos 2θ log(cosθ+sinθ/cosθ-sinθ)=?

7 years ago

Answers : (2)

Anil Pannikar AskiitiansExpert-IITB
85 Points
										

Dear ,


 


cosθ + sinθ = √(1 + sin2θ)


cosθ - sinθ = √(1 - sin2θ)


now inregral becomes,  ∫cos2θ*log√(1 + sin2θ) - ∫cos2θlog√(1 - sin2θ)


put √(1 + sin2θ) = t so, cos2θ dθ = tdt


so, ∫cos2θ*log√(1 + sin2θ) = ∫t*logtdt   ...........1


and,


put √(1 - sin2θ) = X so, -cos2θ dθ = XdX


so, ∫cos2θ*log√(1 + sin2θ) = -∫X*logXdX   ...........2


now solve 1 and 2 by parts...


 


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Askiitians Expert


Anil Pannikar


IIT Bombay

7 years ago
vikas askiitian expert
510 Points
										

I=cos2@log(cos@+sin@/cos@-sin@) d@


 cos@+sin@ can be written as sqrt{(cos@+sin@)^2}=sqrt(1+sin2@)


 cos@-sin@ can be written as sqrt{(cos@-sin@)^2}=sqrt(1-sin2@)


now I becomes integral cos2@logsqrt{(1+sin2@)/sqrt(1-sin2@)}d@                or


             I=COS2@/2 .LOG(1+SIN2@/1-SIN2@)d@


 now put sin2@=t


     so 2cos2@d@=dt


  I=1/4 .log(1+t/1-t)dt


now using integration by parts


I=1/4{  tlog(1+t/1-t) - integral(1-t)t/1+t dt } 


  =1/4 { tlog1+t/1-t  - integral(2-t-1/t+1)dt


  =1/4{  t.log1+t/1-t  - integral( 2 - t  -2/1+t )dt


  =1/4{ tlog1+t/1-t   -  (2t -t^2/2 -2logt+1)  +c     or


 I=1/4 { sin2@2@(log1+sin2@/1-sin2@)  - 2sin2@ +(sin2@)^2 /2 +2log(1+sin2@)  +c


 approve my ans if u like


 


 


 

7 years ago
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