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Community Discussion Question:
integ
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 Nov 2007 22:48:32 IST
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pl inegrate sqrt( x/(x-1))
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 Nov 2007 23:10:26 IST
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Hi, Look, i am not using the integration sign, assume it to be present : sqrt( x/(x-1)) x/(x*(x-1))1/2 2x-1+1/[2(x*(x-1))1/2 ] put x(x-1) = t So, dt = (2x-1) dx So, dt/[2(t)1/2 ] + 1/[2(x*(x-1))1/2 ] leave the second expression as it is now. first can be easily integrated now. For second, amke it a perfect square and then use the special integration formulae to get the final answer
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 Nov 2007 23:44:18 IST
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In ur question initially substitute x=sec2A. then denominator will be tan2A the integral will reduce to (sec2A/tan2A)1/2. but as x=sec2A---------->> dx=( 2secA * secA * tanA)da so the final integral will look like (secA / tanA) * dx which will lead 2 cancellation of tanA resulting in ---------------> 2sec3A dA ------------> now rewrite the same in terms of cosA and next integral will luk like 2/cos3A dA, express this as 2cosA/cos4A dA now express denom as (1-sin2A)*(1-sin2A). Now finally put sinA as ' t ' then cosA dA will be dt and the end result will be 2dt / (1-t2)(1-t2) .Use partial fractions and finish it!!! Hope my detailed steps help u . Rate me if u feel dey helped u.
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