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Integration problems, help!

Prove that,\[\int_{0}^{\infty}\dfrac{1}{x^3}\sin^3\left(x-\dfrac{1}{x} \right)^5\, dx=0\] and\[\int_{-1}^{0}\dfrac{x^2+2x}{\ln{(x+1)}} dx=\ln{3}\].I have no idea how to solve these problems!

Note by Adarsh Kumar
2 years, 5 months ago

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Comment deleted Jun 28, 2015

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Sir, Ur Solution is wrong. Please check it again. There are a lot of mistakes.

Rajdeep Dhingra - 2 years, 5 months ago

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How did you get from the second last step to the last step,or more precisely how did you integrate those functions?

Adarsh Kumar - 2 years, 5 months ago

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@Adarsh Kumar @Rajdeep Dhingra The first Integral is really hard. Also, it is not equal to \(0.\) The Integral is equivalent to \[-\frac15 \int_0^\infty \sin^3(u) u^{-3/5} du \approx {-0.1535} \] Anyway, to be able to compute the original Integral, you need to know some conditions for which an Integral becomes Riemann Improper Integrable. I can try to proceed if you want, but this is really miles and miles above our level.

Ishan Dasgupta Samarendra - 2 years, 5 months ago

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@Adarsh Kumar Seems you've got some excellent answers. I would have done the second the same way as Rajdeep.

Ishan Dasgupta Samarendra - 2 years, 5 months ago

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Sir the 1st one ? @Samarpit Swain solution is wrong.

Rajdeep Dhingra - 2 years, 5 months ago

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By the way really nice solution!

Ishan Dasgupta Samarendra - 2 years, 5 months ago

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@Ishan Dasgupta Samarendra Thank You sir

Rajdeep Dhingra - 2 years, 5 months ago

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Come on Rajdeep! I'm not Sir!\[\]I'll have a go at it in half an hour.

Ishan Dasgupta Samarendra - 2 years, 5 months ago

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@Ishan Dasgupta Samarendra Yes sir U are "sir" for me.

Rajdeep Dhingra - 2 years, 5 months ago

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Comment deleted Jun 29, 2015

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U made terrible mistakes

Kyle Finch - 2 years, 5 months ago

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Sir , I don't see any. Can U please tell ?

Rajdeep Dhingra - 2 years, 5 months ago

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@Rajdeep Dhingra oh sorry, on looking closely it shows \(\frac{1}{a+1}\) otherwise i thought u had written \(\frac{1}{a-1}\)

Kyle Finch - 2 years, 5 months ago

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Wow!Thanx for your solution!You are a genius!

Adarsh Kumar - 2 years, 5 months ago

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Thanks ! I am removing this solution.(as you have seen it)

Rajdeep Dhingra - 2 years, 5 months ago

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Hint about second integral, take x+1=t, and do it with the help of differentiation under integral.

Ronak Agarwal - 2 years, 5 months ago

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Actually i tried that and I got,\[\int_{0}^{1}\dfrac{t^2-1}{\ln{t}}dt\],how to proceed now?

Adarsh Kumar - 2 years, 5 months ago

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Hello Adarsh. I'm really sorry but I won't be able to help now; I'm truly exhausted after coaching class since 10 in the morning - and the same schedule is in place for tomorrow. I'll answer as soon as I can for sure.

Ishan Dasgupta Samarendra - 2 years, 5 months ago

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No problem bhaiya!I understand!

Adarsh Kumar - 2 years, 5 months ago

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