## FoolForMath 3 years ago 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!

1. angela210793

How can integrals be wonderful -_-

2. waheguru

They are once you know them -_=

3. 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...

4. waheguru

Fancy Tricks?

5. FoolForMath

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

6. 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

7. FoolForMath

Changing the bound may work for the first problem :)

8. FoolForMath

9. 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?

10. 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}}$

11. 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...

12. TuringTest

* -2(u-1)

13. 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 $$(1,2)$$.

14. 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...

15. FoolForMath

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

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

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

18. FoolForMath

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

19. Mr.Math

Yeah sure! :)

20. FoolForMath

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

21. Mr.Math

Don't give any any hint!

22. FoolForMath

Lol sorry! :P

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

24. KingGeorge

It doesn't seem so correct anymore...

25. robtobey

3. Refer to the attachment.

26. FoolForMath

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

27. Ishaan94

Oh Sorry.