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Integral Calculus

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11 May 2008 13:04:02 IST

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Challenge in Integration - From Feynmann

None

Solve the following in the shortest possible way

$If\;a>b>0,\;show\;that$ $\int_{0}^{\pi}({sin}^{2}x)/({a}^{2}\;+{b}^{2}-2ab\;cosx) \;dx\;=\frac{\pi}{2{a}^{2}}$

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abhishek sinha

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12 May 2008 19:26:43 IST

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a hint !!

this problem WOULD NOT REQUIRE ANY TYPE OF SUBSTITUTION OR BY PARTS ETC .

THE ULTIMATE INTEGRATION IS JUST A FORMULA AVAILABLE IN THE FIRST PAGE OF ANY INTEGRATION BOOK .

IT JUST REQUIRES A LI'L BIT OF IMAGINATION !!

pink_ele

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14 May 2008 12:28:03 IST

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here a>b>0

let a side X forms triangle wid a and b den

dis gives

a^2+b^2-2abcosx=X^2

ryt

now

(sinx/X)^2=k(constant)................constant=(1/4R^2)

thus given integral is integral of a constant

integral=pi x constant

thats it !!!!!!1

rest ........still m solving

abhishek sinha

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14 May 2008 18:10:01 IST

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You are very close !!

Please complete the remaining steps !!

pink_ele

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14 May 2008 18:20:49 IST

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bhaiyya
u mean to prove R=a/rt 2
fyn......
i'll try:)

abhishek sinha

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Joined: 18 Dec 2007

Posts: 926

15 May 2008 12:27:02 IST

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Now I am providing the complete solution to this problem :

For ease of understansing , let us change the variable x to C

Now consider a triangle whose sides are a, b ( fixed )(a>b) and their intermediate angle C is variable .

Now it is clear that the Dr. of the given integrand =

${a}^2 +{b}^2 -2ab\;cosC =c^2$

so the integral now becomes

$\int_{0}^{\pi}\frac{sin^2C}{c^2}dC$

Now applying Sine rule , we know that

$\frac{sin^2C}{c^2} =\frac{sin^2A}{a^2}$

again we have that $A+B +C=\pi\\so$ dC = - dA -dB

Now let us imagine what would be the limits of A & B as C varies from 0 to $\pi$ ( with a>b ) .

when C->0 , A-> $\pi$ & B->0

again when C-> $\pi$ , A->0 & B->0 { Draw a figure with a>b & observe the above result yourself !! }

so the integral becomes now

$-\int_{\pi}^{0}\frac{sin^2A}{a^2}\;dA \;-\int_{0}^{0}\frac{sin^2A}{a^2}\;dB$

The last integral is clearly zero !!

So the ultimate integral is

$\int_{0}^{\pi}\frac{sin^2A}{a^2}\;dA$

but a is constant ; so the result is

$\int_{0}^{\pi}sin^2A\;dA/a^2$ =

$\frac{\pi}{2{a}^2}$

( hence proved )

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