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Blazing goIITian

Joined: 26 Oct 2006

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24 Feb 2007 12:29:06 IST

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hey kab is it 4th root?

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24 Feb 2007 12:29:38 IST

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Yes.

Blazing goIITian

Joined: 26 Oct 2006

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24 Feb 2007 12:30:57 IST

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okk lemme try this.........

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Joined: 27 Dec 2006

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24 Feb 2007 12:35:52 IST

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dx/(^{[ 4]}

(1+x⁴))

put x=sqroot(tanQ)

dx=1/2sqroot(tanQ)*sec^2Q.dQ

1/2

1/cosQsqroot(sinQ).dQ

Blazing goIITian

Joined: 26 Oct 2006

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24 Feb 2007 12:37:26 IST

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hey i m doin by a diff. method

Blazing goIITian

Joined: 26 Oct 2006

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24 Feb 2007 12:39:52 IST

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we can do it like this also................
takin x common frm the root n tthen mul. num n deno by x^4.......then we get a simple form.....n by puttin 1+x^-4=t^2
we get a form which can b solved easily

Blazing goIITian

Joined: 26 Oct 2006

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24 Feb 2007 12:40:45 IST

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try this................otherwise lemme know if u dont get
ill post my soln

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24 Feb 2007 12:43:17 IST

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Hey Shakir I think cosQ is in the denominator???

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24 Feb 2007 12:45:05 IST

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ya wait i will solve question p[roperly actually i didnt do it on paper

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24 Feb 2007 12:47:07 IST

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I solved it by algebraic substitution and the answer is correct.But if someone can provide any other cool method then I will be grateful!!

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24 Feb 2007 12:54:30 IST

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just starting frm where i left

cool trigo method

dx/(^{[ 4]}

(1+x⁴))

dx/(^{[ 4]}

(1+x⁴))

put x=sqroot(tanQ)

dx=1/2sqroot(tanQ)*sec²Q.dQ

1/2

1/cosQsqroot(sinQ).dQ

multiply and divide by cosQ

1/2

cosQ/cosQ²sqroot(sinQ).dQ

now put sinQ=t

1/2

dt/(1-t²)sqroot(t)

and integrate!!!!

Manasi

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24 Feb 2007 12:57:10 IST

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gud work shakir.... that brings ur 500th altitude.... whrs the party

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24 Feb 2007 13:03:44 IST

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hey KAB even then it is very LONG bcoz
now it is in form 1/Psqroot(Q)
where P is quadratic and Q is linear
so sorry no eXXXXXXXXXXxtra short method here
!

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24 Feb 2007 13:11:48 IST

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Check my method Shakir

Put 1+x⁴=x⁴y⁴

x⁴=1/(y⁴-1)

dx= -y³dy/x³(y⁴-1)²

So given integral is

-y³dy/[x⁴y(y⁴-1)²]= -

y²dy/x⁴(y⁴-1)²

= -

y²dy/(y⁴-1)

So final answer is (1/2)[(1/2)log{(1+y)/(1-y)]-tan^-1y}+c

So which one you think is shorter???

Blazing goIITian

Joined: 26 Oct 2006

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24 Feb 2007 13:13:37 IST

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hey nice one......i m postin my soln....check my also.........its bit lengthy

Blazing goIITian

Joined: 26 Oct 2006

Posts: 530

24 Feb 2007 13:23:58 IST

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_{[ ]}

^{[ ]} dx/^{[ 4]}

1+x⁴

_{[ ]}

^{[ ]} dx/x^{[4 ]}

1+x^-4

_{[ ]}

^{[ ]} x⁴dx/x⁵^{[4 ]}

1+x^-4

putting 1+x^-4=t⁴

-4x^-5dx=4t³dt

_{[ ]}

^{[ ]} t²dt/t⁴ - 1

n it can be solved further..................so howz this one???

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24 Feb 2007 13:27:37 IST

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KAB ingenious!!! method
but i think that even my method is short but not extraaaaaaaaaa short!
and yeah gd method ruhi!!!!!!!!!!!

Blazing goIITian

Joined: 26 Oct 2006

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24 Feb 2007 13:28:43 IST

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hey shakir bhai aise hi bol rahe ho ya sach main theek hai???