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Integral Calculus

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2 Jul 2008 13:03:36 IST

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Integral in terms of another

None

Given that A = $\int_0^{\infty} \frac{\sin x}{x} dx$

find $\int_0^{\infty} \frac{\sin^2 x}{x^2} dx$

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vnkt. swaroop

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Joined: 13 Jun 2008

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2 Jul 2008 22:59:56 IST

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mr.forum expert integral fo sinx/x is not possible.then how can we find integral of Asquare.

ankur gupta

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3 Jul 2008 00:01:03 IST

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i think the answer is 2A
if it is correct then
i wl post the solution .......

Hari Shankar

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3 Jul 2008 09:50:19 IST

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venkat, have you considered that if someone posts a complete solution you may end up looking quite silly?

You cannot even edit your post now!

Hari Shankar

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3 Jul 2008 09:51:16 IST

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venkat, have you considered that if someone posts a complete solution you may end up looking quite silly?

You cannot even edit your post now!

Ankur, your answer is close enough for you to post the solution. Please post it.

ankur gupta

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3 Jul 2008 15:30:36 IST

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sry 4 the wrng typing nw here's the solution

integral (sin²x) / x²dx

nw applying by parts taking 1/x² as one part and sin²x as second part=

= (sin²x) integral (1/x²)dx - integral (- sin2x)dx / x

= [ (-sin²x) / x ]₀^infinity+ integral (sin2x) dx / x

= integral (sin2x)dx / x

nw put 2x = t

2dx = dt

limits remain same

so,= integral 2 (sint dt) / 2 t

= integral (sint dt) / t

= A

thats the answer .......

Hari Shankar

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3 Jul 2008 17:41:36 IST

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Good answer. I wish you would use the editor to make it easy for others to read it.

abhishek sinha

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4 Jul 2008 12:03:31 IST

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anyways , we can actually evaluate A without going beyond syllabus .

In fact integration of {(sin x) / x}^m with the limits given is also possible .

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