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Integral Calculus

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28 Jul 2009 17:12:33 IST

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???(24&dx)/(1+a×cos?x )^2 above is the actual eqn in MS Word. it is actually . Int. dx/(1+a*cos

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???(24&dx)/(1+a×cos?x )^2 above is the actual eqn in MS Word.it is actually . Int. dx/(1+a*cosx)^2

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kabi

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Joined: 11 Jan 2009

Posts: 575

1 Aug 2009 12:58:22 IST

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Let P = sin(x) / (1 + acos(x))

p ' = ( a + cos (x)) / ( 1 + acos(x))^2

after rearranging it

it will become -

P' = ( a²-1 ) / ( a ( 1+acos(x))² ) + 1 / ( a ( 1+ acos(x))

( a²-1 ) / ( a ( 1+acos(x))² ) =- 1 / ( a ( 1+cos(x)) ) + P '

now integrating this will give ::

( a² -1 ) /a ( integration ( dx / ( 1 +acos(x))²) = integration ( P' dx ) - integration ( dx / ( a (acos(x)+1))

( integration ( dx / ( 1 +acos(x))²) = a / ( a²- 1 ) *P - 1 / ( a²-1) integration ( dx / ( acos (x ) +1 )) ......( 1 )

now for in integration of dx / ( acos(x) +1 )

put cos (x) = ( 1 -tan²x/2 )/ ( 1+tan²x/2 ) then put tan (x/2 ) = t

so integration of dx / ( acos(x) +1 )= integration ( 2dt / ( 1+a + t² ( 1-a ))

which i think u can find .

then just the value of P and above inegral value in equation (1)

thank u

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