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∫cos 2θ log(cosθ+sinθ/cosθ-sinθ)=?

∫cos 2θ log(cosθ+sinθ/cosθ-sinθ)=?

Grade:12

2 Answers

Anil Pannikar AskiitiansExpert-IITB
85 Points
10 years ago

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

vikas askiitian expert
509 Points
10 years ago

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

 

 

 

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