## FoolForMath Group Title 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! 2 years ago 2 years ago

1. angela210793 Group Title

How can integrals be wonderful -_-

2. waheguru Group Title

They are once you know them -_=

3. TuringTest Group Title

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

4. waheguru Group Title

Fancy Tricks?

5. FoolForMath Group Title

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

6. TuringTest Group Title

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

7. FoolForMath Group Title

Changing the bound may work for the first problem :)

8. FoolForMath Group Title

9. TuringTest Group Title

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?

10. TuringTest Group Title

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}}$

11. TuringTest Group Title

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...

12. TuringTest Group Title

* -2(u-1)

13. Mr.Math Group Title

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 $$(1,2)$$.

14. TuringTest Group Title

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...

15. FoolForMath Group Title

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).

16. Mr.Math Group Title

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

17. Mr.Math Group Title

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 $$\frac{3}{2}$$.

18. FoolForMath Group Title

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

19. Mr.Math Group Title

Yeah sure! :)

20. FoolForMath Group Title

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

21. Mr.Math Group Title

Don't give any any hint!

22. FoolForMath Group Title

Lol sorry! :P

23. KingGeorge Group Title

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

24. KingGeorge Group Title

It doesn't seem so correct anymore...

25. robtobey Group Title

3. Refer to the attachment.

26. FoolForMath Group Title

@Ishaan94: Typical huh :P It was supposed to be for non-Indians. REF: http://openstudy.com/users/turingtest#/updates/4f81cd50e4b0505bf0836887

27. Ishaan94 Group Title

Oh Sorry.