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 2 years ago
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
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!

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

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

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

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0I'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

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

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0I 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?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0so 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\[dx={du\over(2uu^2)^{1/2}}\]

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

Mr.Math
 2 years ago
Best ResponseYou've already chosen the best response.3I 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 \((1,2)\).

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0my 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...

FoolForMath
 2 years ago
Best ResponseYou've already chosen the best response.7There 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).

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

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

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

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

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

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.0Want 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 \(\cos x =\sqrt{u^4+2u^2}\)

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

robtobey
 2 years ago
Best ResponseYou've already chosen the best response.03. Refer to the attachment.

FoolForMath
 2 years ago
Best 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
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