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Integral Calculus

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24 Jan 2007 14:47:14 IST

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WHO CAN SOLVE THIS PART2

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Evaluate (tan^-1x)dx/(1+x)

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nrki99

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Joined: 20 Jan 2007

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24 Jan 2007 21:05:56 IST

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hi kab

(tan^-1x)dx/(1+x)

then multiply by 1+x^2 in both num and denominator

(tan^-1x) 1+x^2dx/(1+x) 1+x^2

then let (tan^-1x)=t

then differentiate it u get 1*dx/1+x^2 =dt

dt * 1+tan^2/ 1+tant

sec^2 t/1+tant

then u again let 1+tant=u

and differentiate it u get

sec^2 t dt=dt

du/u =ln[u] [] denotes mod function

so answer becomes ln[1+tant]

ln[1+tan(tan^-1x)]=ln(1+x) ------- ans

CyBorG

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24 Jan 2007 21:35:09 IST

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You forgot to include t in the Nr!!!!!!!

nishant dash

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24 Jan 2007 22:38:16 IST

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exactly.a simple but glaring mistake..

nishant dash

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26 Jan 2007 16:52:20 IST

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could u plz tell how to solve it.i have tried it but didn't get any answer..

$h0cK3r

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26 Jan 2007 17:03:45 IST

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i have seen this question in definte integration we get
integration by parts
tan-1x log(1+x) -integral of( log(1+x)/(1+x^2))
the integral becomes easy by putting x=tan Q
and we get integral of log(1+tanQ) and then we get
integral of log (sin Q+cosQ)-log(cos Q)
we can do this in defnite were we use properties whemn integrated from 0 to pi/2 to prove it equal to -pi/2*log 2

CyBorG

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27 Jan 2007 17:03:31 IST

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Try this

_{[0 ]}^{[ 1]} (tan^-1x)dx/(1+x)

$h0cK3r

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27 Jan 2007 19:24:06 IST

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Integrating by parts taking tan^-1x and 1/1+x as second we get

pi/4 log2 - _{[0 ]}

^{[ 1]} log(1+x)/(1+x²)

put x=tanQ

dx=sec²Q dQ

the term after the - sign gets reduced to _{[ 0]}^{[ pi/4]}log(1+tanQ)

which can be integrated using properties easily and is a standard NCERT problem

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