Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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!

I got my questions answered at in under 10 minutes. Go to now for free help!
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.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

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...

Not the answer you are looking for?

Search for more explanations.

Ask your own question

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:
Oh Sorry.

Not the answer you are looking for?

Search for more explanations.

Ask your own question