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..
Steve
answered
√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
Shenaya
answeredI thought they were expecting a whole number!
Steve
answered
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...
Steve
answered
∫[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.
Explain Bot
answered
Start with the integral expression:
I = ∫[x * sqrt((a^2 - x^2)/(a^2 + x^2))] dx from 0 to a
Using the substitution u = sqrt(x^2 + a^2), we have:
I = ∫[sqrt(2a^2 - u^2)] du from a to sqrt(2a^2)
Now, let's follow the steps you mentioned:
1. Substitute (u / sqrt(2a^2)) = sin(t), which leads to du = sqrt(2a^2) * cos(t) dt.
2. Substitute the limits u = a and u = sqrt(2a^2) into the new integral expression.
After applying these substitutions and simplifying, we have:
I = sqrt(2a^2) * ∫[sqrt(1 - sin^2(t))] cos(t) dt from π/6 to π/4
I = sqrt(2a^2) * ∫[cos^2(t)] dt from π/6 to π/4
Now, integrating cos^2(t) over the given limits, we get:
I = sqrt(2a^2) * [(sin(2t) / 2) + t] evaluated from π/6 to π/4
I = sqrt(2a^2) * [(1/2 - sqrt(3)/4) + (π/4 - π/6)]
I = sqrt(2a^2) * [(1/2 - sqrt(3)/4) + (π/12)]
I = sqrt(2a^2) * [(6 - 3sqrt(3) + π) / 12]
There doesn't seem to be any mistake in your calculations so far. However, we need to simplify this expression and compare it with (π - 2) to find the largest value of 'a'.
To do this, we can simplify the expression further as follows:
I = (sqrt(2a^2) / 12) * (6 - 3sqrt(3) + π)
Now, compare this with (π - 2):
I = (sqrt(2a^2) / 12) * (6 - 3sqrt(3) + π) ≤ (π - 2)
We are trying to find the greatest value of 'a' such that this inequality holds.
Comparing the coefficients of sqrt(2a^2) on both sides, we have:
(sqrt(2) / 12) * (6 - 3sqrt(3) + π) ≤ 1
Now, we can solve this inequality for 'a' by isolating it:
(sqrt(2) / 12) * (6 - 3sqrt(3) + π) ≤ 1
(sqrt(2) / 12) * (6 - 3sqrt(3) + π) * 12 / sqrt(2) ≤ 1 * 12 / sqrt(2)
6 - 3sqrt(3) + π ≤ 12 / sqrt(2)
Simplifying further:
6 - 3sqrt(3) + π ≤ 6sqrt(2)
Rearranging terms:
3sqrt(3) + π ≥ 6sqrt(2) - 6
Now, we can solve this inequality to find the critical value of π:
3sqrt(3) ≥ 6sqrt(2) - 6
sqrt(3) ≥ 2sqrt(2) - 2
Squaring both sides:
3 ≥ 8 - 4sqrt(2) + 4
4sqrt(2) ≥ 9
sqrt(2) ≥ 9/4
This means that for the value of π greater than 9/4, the inequality holds true. Hence, the largest value of 'a' would be for π = 9/4.
Therefore, the greatest value of 'a' satisfying the integral expression is when π = 9/4.
I hope this explanation helps you identify any mistakes you made and understand the correct approach to solve this problem.
