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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 non-zero 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!

Mathematics
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How can integrals be wonderful -_-
They are once you know them -_=
I don't know those fancy tricks they seem to teach you guys in India, so I'll have to try to be creative...

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Other answers:

Fancy Tricks?
@TuringTest: The first 2 are somewhat standard trick. The other two require NOT so standard trick ;)
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
Changing the bound may work for the first problem :)
Added the answer, so that you guys can check your result.
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?
so far for the first I have\[\int{\sin x\over\sqrt{\sin x+1}}dx\]\[u=1+\sin x\implies \sin x=u-1\]\[du=\cos xdx=(1-\sin^2x)^{1/2}dx=(1-(u-1)^2)^{1/2}dx=(2u-u^2)^{1/2}dx\]therefor\[dx={du\over(2u-u^2)^{1/2}}\]
so\[\int{u-1\over u^{1/2}(2u-u^2)^{1/2}}du\]and I'm thinking by parts since \[(2u-u^2)'=-2(1-u)\]let's se...
* -2(u-1)
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: \(f'(\alpha)=\frac{f(2)-f(1)}{2-1}\). 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 \((1,2)\).
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...
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 at-least one real root in (a,b).
This property is nothing but a direct corollary of the mean value theorem.
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 \(\frac{3}{2}\).
MrMath I agree, but in competitive it helps though :)
Yeah sure! :)
Although, trivial lol! For the fourth one there is a inequality which you can use.
Don't give any any hint!
Lol sorry! :P
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 \(\sin x =u^2-1\). From this, we get that \(\sin^2 x = u^4-2u^2+1\). This means that \(\cos^2(x) = -u^4+2u^2\) so \(\cos x =\sqrt{-u^4+2u^2}\)
It doesn't seem so correct anymore...
3. Refer to the attachment.
@Ishaan94: Typical huh :P It was supposed to be for non-Indians. REF: http://openstudy.com/users/turingtest#/updates/4f81cd50e4b0505bf0836887
Oh Sorry.

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