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Integral Calculus

astha jain's Avatar
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13 Mar 2010 08:30:51 IST
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prove that... integration x / (1-cos a sinx) wrt x within limits 0 to pie = pie(pie-a
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prove that... integration x / (1-cos a sinx) wrt x within limits 0 to pie = pie(pie-a)/sin a plzzz reply


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Deepak Aggarwal's Avatar

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Joined: 9 May 2009
Posts: 766
13 Mar 2010 11:52:33 IST
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I got a different answer to it.

I am getting I = (pi)2 / 2 sin(a).

nyhow approach is like dis.

First apply property int x from 0 to a [ f(x) ] = int x from 0 to a [ f(a - x) ] and then apply

int x from 0 to 2a [ f(x) ] = 2 int x from 0 to a [ f(x) ] , if f(2a - x) = f(x)

U'll get,

I = (pi) int x from 0 to (pi/2) ( 1 / 1 - cos(a) sin(x) ).dx

Let I1 = int x from 0 to (pi/2) ( 1 / 1 - cos(a) sin(x) ).dx

Now put cos(a) sin(x) = y and u'll get

I1 = int y from 0 to cos(a) ( 1 / (1 - y) (cos2a - y2)1/2 ).dy

Now put cos2a - y2 = 1/t

U will now get, I1 = (1/sina) int t from 1 to (1 / 1 - cos(a)) [ (cos(a)/sin2(a))2 - (t - (1/sin2a))2 ]1/2

So, finally I1 = pi / 2 sin(a)

So, I = (pi)2 / 2 sin(a) [Ans]..............

 

 

 

 




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