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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: How to do this integration?
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[Post New]23 Jun 2008 19:30:45 IST
    Subject: How to do this integration?
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juana (44)

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root of cot x dx?
    
[Post New]23 Jun 2008 19:57:15 IST
    Subject: Re:How to do this integration?
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sboosy (3046)

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 \int \sqrt{\cot(x)} \,dx \\ \\ \cot(x)=t^2 \ -\csc^2(x)dx = 2tdt \\ \\ \Rightarrow dx=\frac{2tdt}{-(1+t^4)} \\ \\ \mbox{So integral reduces to} \\ \\ - \int \frac{t^2+1}{1+t^4} \,dt - \int \frac{t^2-1}{1+t^4} \,dt \\ \\ = - \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}} \,dt - \int \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}} \,dt \\ \\ \mbox{First integral,put} \ t-\frac{1}{t} = p , \ \mbox{Second,put} \ t+\frac{1}{t}=q \\ \\ \mbox{So they reduce to} \ - \int \frac{1}{p^2+2} \,dp - \int \frac{1}{q^2-2} \,dq \\ \\ \mbox{which are 2 integrals in standard form ..thus u can proceed}
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[Post New]23 Jun 2008 20:39:11 IST
    Subject: How to do this integration?
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juana (44)

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how did the second step come? what is t square? is the second line 2 diff steps?
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[Post New]23 Jun 2008 21:00:38 IST
    Subject: How to do this integration?
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paddy.dude (1152)

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yup second line is 2 diff step
in the first he has put cotx as equal to t and then he has differentiated it in the second step
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[Post New]25 Jun 2008 22:28:45 IST
    Subject: How to do this integration?
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juana (44)

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ok thanks
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[Post New]25 Jun 2008 22:56:10 IST
    Subject: How to do this integration?
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juana (44)

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sboosy hoe did you do after assuming p and q?
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[Post New]25 Jun 2008 23:03:01 IST
    Subject: Re:How to do this integration?
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paddy.dude (1152)

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he modified t2+1/t2 as ( t-1/t)2+ 2


and the other one as (t+1/t)2 -2


now he put   t-1/t=p


which on differentiating


(1+1/t2) dx=dp


which is  the numerator


similarly for the other one also


then its a std integral
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