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FoolForMath
Group Title
Fool's problem of the day,
Wonderful Integrals:
1. Evaluate: \( \huge \int \frac{\sin x}{\sqrt{1+\sin x}} \)
2. Prove \( \huge \int \limits_{0}^{\frac \pi 2}\frac{x}{\sin x+\cos x}dx = \frac{\pi}{2\sqrt{2}} \ln (\sqrt{2} +1) \)
3. Prove \( \int \limits_{0} ^1 \sqrt{(1+x)(1+x^3)} \; dx \le \sqrt{\frac{15}{8}} \)
4. Let \( a,b, c \) be nonzero real numbers, such that \( \int \limits_{0} ^1 (1+ \cos ^8 x)(ax^2 bx +c) \; dx = \int \limits_{0} ^2 (1+ \cos ^8 x)(ax^2 bx +c) \; dx \). Then show that the quadratic equation \( ax^2 +bx+c=0\) has one root in \( (1,2) \)
[Solved by @Mr.Math]
PS: First two are probably very easy! ;)
@TuringTest Time for you solve my problem of the day! Between I would only accept all the solutions from you ;)
Good luck!
 2 years ago
 2 years ago
FoolForMath Group Title
Fool's problem of the day, Wonderful Integrals: 1. Evaluate: \( \huge \int \frac{\sin x}{\sqrt{1+\sin x}} \) 2. Prove \( \huge \int \limits_{0}^{\frac \pi 2}\frac{x}{\sin x+\cos x}dx = \frac{\pi}{2\sqrt{2}} \ln (\sqrt{2} +1) \) 3. Prove \( \int \limits_{0} ^1 \sqrt{(1+x)(1+x^3)} \; dx \le \sqrt{\frac{15}{8}} \) 4. Let \( a,b, c \) be nonzero real numbers, such that \( \int \limits_{0} ^1 (1+ \cos ^8 x)(ax^2 bx +c) \; dx = \int \limits_{0} ^2 (1+ \cos ^8 x)(ax^2 bx +c) \; dx \). Then show that the quadratic equation \( ax^2 +bx+c=0\) has one root in \( (1,2) \) [Solved by @Mr.Math] PS: First two are probably very easy! ;) @TuringTest Time for you solve my problem of the day! Between I would only accept all the solutions from you ;) Good luck!
 2 years ago
 2 years ago

This Question is Closed

angela210793 Group TitleBest ResponseYou've already chosen the best response.0
How can integrals be wonderful _
 2 years ago

waheguru Group TitleBest ResponseYou've already chosen the best response.0
They are once you know them _=
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I don't know those fancy tricks they seem to teach you guys in India, so I'll have to try to be creative...
 2 years ago

waheguru Group TitleBest ResponseYou've already chosen the best response.0
Fancy Tricks?
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.7
@TuringTest: The first 2 are somewhat standard trick. The other two require NOT so standard trick ;)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I'm just getting used to the ideas of changing bounds in problems like these, so I'm clumsy with that trick. It is my initial idea though
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.7
Changing the bound may work for the first problem :)
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.7
Added the answer, so that you guys can check your result.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I feel I have a good start on the first one but I suddenly can't see the latex is everyone else experiencing the site slowing down?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
so far for the first I have\[\int{\sin x\over\sqrt{\sin x+1}}dx\]\[u=1+\sin x\implies \sin x=u1\]\[du=\cos xdx=(1\sin^2x)^{1/2}dx=(1(u1)^2)^{1/2}dx=(2uu^2)^{1/2}dx\]therefor\[dx={du\over(2uu^2)^{1/2}}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
so\[\int{u1\over u^{1/2}(2uu^2)^{1/2}}du\]and I'm thinking by parts since \[(2uu^2)'=2(1u)\]let's se...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
* 2(u1)
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.3
I will do the fourth problem. Let \(f(x)=\int_0^x (1+\cos^8(t))(at^2+bt+c)dt\). Then, by the mean value theorem, there exists a point \(\alpha\) in \((1,2)\) such that: \(f'(\alpha)=\frac{f(2)f(1)}{21}\). But \(f(2)=f(1) \implies f'(\alpha)=0\). Thus \((1+\cos^8(\alpha))(a\alpha^2+b\alpha+c)=0 \implies a\alpha^2+b\alpha+c=0\), since \(1+\cos^8(\alpha)\) has no real zeros in \((1,2)\).
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
my method seems not to work, and I have to go again I am completely foiled by your problems FFM, though they are on one of my favorite topics :( some day though, some day...
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.7
There is property which makes things much easy for #4. If \( f(x) \) is continuous on [a,b] and \( \int \limits_{a}^{b} f(c)=0\); then equation f(x)=0 has atleast one real root in (a,b).
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.3
This property is nothing but a direct corollary of the mean value theorem.
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.3
For the third problem, I haven't yet found the boundary you gave for the integral. The least upper bound I found so far is \(\frac{3}{2}\).
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.7
MrMath I agree, but in competitive it helps though :)
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.3
Yeah sure! :)
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.7
Although, trivial lol! For the fourth one there is a inequality which you can use.
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.3
Don't give any any hint!
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.7
Lol sorry! :P
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Want to solve\[\int\limits {\sin x \over \sqrt{1+\sin x}}\quad dx\]Let \(u=\sqrt{1+\sin x}\). Then we have the relation \(\sin x =u^21\). From this, we get that \(\sin^2 x = u^42u^2+1\). This means that \(\cos^2(x) = u^4+2u^2\) so \(\cos x =\sqrt{u^4+2u^2}\)
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
It doesn't seem so correct anymore...
 2 years ago

robtobey Group TitleBest ResponseYou've already chosen the best response.0
3. Refer to the attachment.
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.7
@Ishaan94: Typical huh :P It was supposed to be for nonIndians. REF: http://openstudy.com/users/turingtest#/updates/4f81cd50e4b0505bf0836887
 2 years ago
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