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`        ∫cos 2θ log(cosθ+sinθ/cosθ-sinθ)=?`
7 years ago

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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Anil Pannikar
IIT Bombay
```
7 years ago
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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