plz solve the sum

Asked by prabhat1993 | 10th Feb, 2010, 09:35: PM

Expert Answer:

Dear Student

I = S ln(1-cosx)

I = S ln(1+cosx)

2I=S ln(sin^2x)

I = S lnsinx

Now solve this

S---->Integral sign

I=  S o to pi   log (sin x)) dx

= 2 S o to pi/2    log (sin x)) dx-----------(1)

=S o to pi /2   log (sin x)) dx+  S o to pi/2   log (cos x)) dx

=S o tp pi/2   log (sin  2x)) dx-S o to pi/2 log 2  dx

=(1/2)  S o tp pi   log (sin t)) dt-pi/2  log 2

=S o tp pi/2   log (sin t)) dt-pi/2  log 2

 

Consider

2 S o to pi/2   log (sin t) dt = S o to pi/2   log (sin t) dt-pi/2 log 2

S o to pi/2   log (sin t) dt= -pi/2     log2

=====>  S o to pi/2   log (sin t) dt= -pi log 2

I= -pi log 2 -pi/2   log2 = -3pi/2     log 2

Regards

Teram

Topperlearning.com

 

 

Answered by  | 13th Feb, 2010, 11:53: AM

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