Find the greatest value of a,so that

Question

integrate [x*root{(a^2-x^2)/(a^2+x^2)} ] from 0-a<=(π-2)

Let I =integrate{ [x*root{(a^2 - x^2)/(a^2+x^2)}]dx } from 0-a

I used the substitution u=root[x^2+a^] to simplify the further.

After substituting and changing limits I got,

I=integrate.{root(2a^2 - u^2) du } from a-√2a

Then I used the substitution (u/2a)=sin t

And after simplifications and changing limits I got,

I=integrate{ √2(a)^2 *2(cos t)^2 dt} from (π/6)-(π/4)

Then finally I got,

I= √2(a)^2[ {(sin 2t)/2} + {t} ] ,limits π/6-π/4

I= √2(a)^2{ [ 1/2 - √3/4 ] + [π/4 - π/6 ]
I=√2(a)^2{ [(2-√3)/4] + [π/12] }

I=√2(a)^2{ [6-3√3+π]/12 }

But I get the feeling that I've made a mistake somewhere,because I don't see any way to simply this further so that it can be compared with (π-2) to find the largest value of a.But can't find where I've made the mistake..

Answer 1

well, just looking at the calculations, they appear reasonable, so at the end you have

√2 a^2 [6-3√3+π]/12 <= π-2
a^2 <= 12(π-2)/(√2 [6-3√3+π])

|a| <= √(12(π-2)/(√2 [6-3√3+π])) = 1.567

Answer 2

Did you find any mistakes in my calculations?

I thought they were expecting a whole number!

Answer 3

Ok, let me check on that

u^2 = x^2+a^2
2u du = 2x dx
a^2-x^2 = 2a^2-u^2

and now
∫[0,a] x√((a^2-x^2)/(a^2+x^2)) dx
= ∫[a,a√2] 1/u √(2a^2-u^2) du

You seem to have lost the factor of 1/u

Now let
u=√2 a sint
2a^2-u^2 = 2a^2-2a^2sin^2t = 2a^2 cos^2t
du = √2 a cost dt
and the integral is now
∫[a,a√2]1/u √(2a^2-u^2) du
=∫[π/6,π/4] 1/(√2 a sint)*√2 a cost * √2 a cost dt
You seem to have lost your du->dt conversion
= ∫[π/6,π/4] √2 a sint cos^2t dt
= -a√2/3 cos^3t [π/6,π/4]
= -a√2/3 ((1/√2)^3-(√3/2)^3)
= -a√2/3 (1/(2√2)-3√3/8)
= a/3 (3√6-4)

Still not an integer!
As always, double-check my math...

Answer 4

Oops. I see a big one! That 1/u factor makes it

∫[π/6,π/4] √2 a cos^2t/sint dt
a√2 ∫[π/6,π/4] (1-sin^2t)/sint dt
a√2 ∫[π/6,π/4] csct-sint dt
a√2 (-ln(csc t + cot t) + cost) [π/6,π/4]
= a√2 (-ln(√2+1)+1/√2)-(-ln(2/√3 + √3)+√3/2))
= a√2 (1/√2+√3/2 + ln(5/(√3+√6)))

Still nowhere near an integer.