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[Post New]10 Jul 2008 10:06:41 IST
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kane (2201)

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the value of the integral


(cos3x+cos5x)/(sin2x+sin4x)dx
there are numerous options besides I.I.T

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[Post New]10 Jul 2008 11:01:31 IST
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sachinguptaiit (940)

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\int\frac{\cos x((1-\sin^2x)+(1-\sin^2x)^2)}{\sin^2x+\sin^4x}.dx\\\\\int\frac{(1-t^2)+(1-t^2)^2}{t^2+t^4}.dt\\\\\int\frac{t^4-3t^2+2}{t^2+t^4}.dt\\\\\int\frac{(t^4+t^2-4t^2+2)}{t^2+t^4}.dt\\\\\int dt-\int\frac{4t^2-2}{t^2(t^2+1)}.dt\\\\t+\int\frac{2}{t^2}.dt-\int\frac{6}{t^2+1}.dt\\\\\sin x -\frac{2}{\sin x}-6\tan^{-1}(\sin x)+ C


 


 
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[Post New]11 Jul 2008 09:48:39 IST
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kane (2201)

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thnxs a lot yaar.what a gr8 way to attempt this question
there are numerous options besides I.I.T

<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>


<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>









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[Post New]13 Jul 2008 17:38:07 IST
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ankitagg (314)

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another way


get cos^3x and sin^2x common


then put sinx=t


cosxdx=dt


dx=dt/sinx


I=int. (1-t^2)(2-t^2)/t^2(1+t^2)


solve by partial frac.


ans.. sinx-2cosecx-6arctan(sinx)+C


 
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[Post New]13 Jul 2008 18:07:43 IST
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venkat_tatolu (228)

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Integrand is the form of F(sinx,cosx)

1)If F(-Sinx,Cosx)=- F(Sinx,Cosx) then put cosx=t

2)If F(Sinx,-Cosx) =- F(Sinx,CosX) then put sinx=t.

3) If F(-sinx,-cosx)= F(sinx,cosx) then put tanx=t.

If 1) point failed then go to second point.If second point failed go to 3rd point.
Given problem is of second type.So put sinx=t.
If u remember above points then u can do any problem of above type.All the best.
***T.Venkat***
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[Post New]14 Jul 2008 22:31:06 IST
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kane (2201)

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thnxs guys
there are numerous options besides I.I.T

<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>


<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>









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