Question

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!

Answer 1

How can integrals be wonderful -_-

Answer 2

They are once you know them -_=

Answer 3

I don't know those fancy tricks they seem to teach you guys in India, so I'll have to try to be creative...

Answer 4

Fancy Tricks?

Answer 5

@TuringTest: The first 2 are somewhat standard trick. The other two require NOT so standard trick ;)

Answer 6

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

Answer 7

Changing the bound may work for the first problem :)

Answer 8

Added the answer, so that you guys can check your result.

Answer 9

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?

Answer 10

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\[dx={du\over(2u-u^2)^{1/2}}\]

Answer 11

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...

Answer 12

*
-2(u-1)

Answer 13

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 \((1,2)\).

Answer 14

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...

Answer 15

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).

Answer 16

This property is nothing but a direct corollary of the mean value theorem.

Answer 17

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 \(\frac{3}{2}\).

Answer 18

MrMath I agree, but in competitive it helps though :)

Answer 19

Yeah sure! :)

Answer 20

Although, trivial lol! For the fourth one there is a inequality which you can use.

Answer 21

Don't give any any hint!

Answer 22

Lol sorry! :P

Answer 23

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 \(\cos x =\sqrt{-u^4+2u^2}\)

Answer 24

It doesn't seem so correct anymore...

Answer 25

3.
Refer to the attachment.

Answer 26

@Ishaan94: Typical huh :P It was supposed to be for non-Indians.

REF: http://openstudy.com/users/turingtest#/updates/4f81cd50e4b0505bf0836887

Answer 27

Oh Sorry.