anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
angela210793
  • angela210793
How can integrals be wonderful -_-
waheguru
  • waheguru
They are once you know them -_=
TuringTest
  • TuringTest
I don't know those fancy tricks they seem to teach you guys in India, so I'll have to try to be creative...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

waheguru
  • waheguru
Fancy Tricks?
anonymous
  • anonymous
@TuringTest: The first 2 are somewhat standard trick. The other two require NOT so standard trick ;)
TuringTest
  • TuringTest
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
anonymous
  • anonymous
Changing the bound may work for the first problem :)
anonymous
  • anonymous
Added the answer, so that you guys can check your result.
TuringTest
  • TuringTest
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?
TuringTest
  • TuringTest
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\[dx={du\over(2u-u^2)^{1/2}}\]
TuringTest
  • TuringTest
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...
TuringTest
  • TuringTest
* -2(u-1)
Mr.Math
  • Mr.Math
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 \((1,2)\).
TuringTest
  • TuringTest
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...
anonymous
  • anonymous
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).
Mr.Math
  • Mr.Math
This property is nothing but a direct corollary of the mean value theorem.
Mr.Math
  • Mr.Math
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 \(\frac{3}{2}\).
anonymous
  • anonymous
MrMath I agree, but in competitive it helps though :)
Mr.Math
  • Mr.Math
Yeah sure! :)
anonymous
  • anonymous
Although, trivial lol! For the fourth one there is a inequality which you can use.
Mr.Math
  • Mr.Math
Don't give any any hint!
anonymous
  • anonymous
Lol sorry! :P
KingGeorge
  • KingGeorge
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 \(\cos x =\sqrt{-u^4+2u^2}\)
KingGeorge
  • KingGeorge
It doesn't seem so correct anymore...
anonymous
  • anonymous
3. Refer to the attachment.
anonymous
  • anonymous
@Ishaan94: Typical huh :P It was supposed to be for non-Indians. REF: http://openstudy.com/users/turingtest#/updates/4f81cd50e4b0505bf0836887
anonymous
  • anonymous
Oh Sorry.

Looking for something else?

Not the answer you are looking for? Search for more explanations.