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Discussion Response Post to: DEFINITE INTEGRAL

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puneet (2841)

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hiiiiiiii liku

this is typically a very different kind of question ... when u read the solution u may wonder that what is going on .. but just relax .... look at question .. the first idea that shud strike u is that what shud i use ..

well clearly no property seems to be applicable here ... but there is some hope .. there is an arbitary m in the expression to be integrated ..

this gives an idea .. we can use the reduction formula ..

so let us try our hand on it and see what happens ..

so let us call I_m = ₀

sin²mx/sin²x dx

= ₀

(1-cos2mx)/(1-cos2x) dx

Now put 2x = t

So now I_m = 1/2.₀

² (1-cosmt)/(1-cost) dt

= 1/2.2 ₀

(1-cosmt)/(1-cost) dt

= ₀

(1-cosmt)/(1-cost) dt

So now I_m+1 - I_m = ₀

[(1-cos(m+1)t)/(1-cost) - (1-cosmt)/(1-cost)] dt

= ₀

(cosmt -cos(m+1)t)/(1-cost) dt

= ₀

(2.sint/2.sin(m+1/2)t)/(2sin²t/2) dt

so, I_m+1 - I_m = ₀ sin(m+1/2)t/sin(t/2) dt

similarily I_m - I_m-1 = ₀ sin(m-1/2)t/sin(t/2) dt

so (I_m+1 - I_m) - (I_m - I_m-1) = ₀[sin(m+1/2)t - sin(m-1/2)t]/sin(t/2)dt

= ₀[2cosmt.sin(t/2)]/sin(t/2)dt

= 0

so (I_m+1 - I_m) - (I_m - I_m-1) = 0

or, 2.I_m = I_m+1+ I_m-1

or, I_m-1,I_m,I_m+1are in A.P.

Now, I_m = ₀

sin²mx/sin²x dx

thus, I₀ = ₀

sin²0x/sin²x dx

= 0

and, I₁ = ₀ sin²1x/sin²x dx

= ₀ dx =

So common difference of A.P is

and hence I_m = I₀ + n.

= n.

... ans

I hope u got this tough one .. :)

cheers

Puneet Agrawal IIT Delhi

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