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The Proof is Trivial!

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Original post by Benjy100
Problem 33

Determine ∫ba1(a−x)(x−b)dx\int\limits_b^a {\frac{1}{{\sqrt {(a - x)(x - b)} }}} {\rm{ }}dx


Solution 33

∫ba1(a−x)(x−b)dx=∫badx(a−b2)2−(x−a+b2)2\displaystyle \int\limits_b^a {\frac{1}{{\sqrt {(a - x)(x - b)} }}} {\rm{ }}dx= \int^a_b \frac{dx}{\sqrt{ \left( \frac{a-b}{2} \right) ^2 - \left( x-\frac{a+b}{2} \right) ^2 }} - upon expanding, completing the square and simplifying.

Now we could use a trig. substitution, but recognise that this is just a standard integral and so

=[arcsin⁡(2x−a−ba−b)]ba\displaystyle = \left[ \arcsin \left( \frac{2x-a-b}{a-b} \right) \right]_b^a

=arcsin⁡(a−ba−b)−arcsin⁡(b−aa−b)\displaystyle = \arcsin \left( \frac{a-b}{a-b} \right) - \arcsin \left( \frac{b-a}{a-b} \right)

=arcsin⁡1−arcsin⁡(−1)=2arcsin⁡1\displaystyle = \arcsin 1 - \arcsin (-1) = 2\arcsin 1 since arcsin is an odd function.

=2×π2=π\displaystyle = 2 \times \frac{\pi }{2} = \pi
(edited 11 years ago)
Original post by Lord of the Flies
Solution 33

Method 1:

Letting x=bcos⁡2t+asin⁡2tx=b\cos^2 t+a\sin^2 t gives:

∫badx(a−x)(x−b)=∫0π/22dt=π\displaystyle \int_b^a \frac{dx}{\sqrt{(a-x)(x-b)}} = \int_0^{\pi/2} 2 dt =\pi

Method 2:

∫ba1(a−x)(x−b)  dx=i∫ba1(a−x)(b−x)  dx\displaystyle\int_b^a \frac{1}{\sqrt{(a-x)(x-b)}}\;dx=i\int_b^a \frac{1}{\sqrt{(a-x)(b-x)}}\;dx


=−2i∫ba(−12a−x+−12b−x)(1a−x+b−x)  dx=\displaystyle -2i\int_b^a\left(\frac{-1}{2\sqrt{a-x}}+\frac{-1}{2\sqrt{b-x}}\right)\left(\frac{1}{\sqrt{a-x}+\sqrt{b-x}}\right)\;dx


=−2iln⁡(a−x+b−x)∣ba=−iln⁡(a−bb−a)=π=-2i\ln(\sqrt{a-x}+\sqrt{b-x})|_b^a=-i\ln\left(\dfrac{a-b}{b-a}\right)=\pi

(taking the principal branch - I'm sure that can be justified)


Too quick! :tongue:
Solution 33
∫ba1(x−b)(a−x)dx=∫ba114(a−b)2−[x−12(a+b)]2dx\displaystyle\int_{b}^{a} \dfrac{1}{\sqrt{(x-b)(a-x)}} dx = \displaystyle\int_{b}^{a} \dfrac{1}{ \sqrt{ \frac{1}{4} (a-b)^{2} - [x-\frac{1}{2}(a+b)]^{2}}} dx

Letting x−12(a+b)=12(a−b)sin⁡θx - \frac{1}{2} (a+b) = \frac{1}{2} (a-b) \sin \theta we get

∫ba114(a−b)2−[x−12(a+b)]2dx=∫−π/2π/2dθ=π\displaystyle\int_{b}^{a} \dfrac{1}{ \sqrt{ \frac{1}{4} (a-b)^{2} - [x-\frac{1}{2}(a+b)]^{2}}} dx = \displaystyle\int_{-\pi /2}^{\pi /2} d \theta = \pi
(edited 10 years ago)
Reply 143
Avatar for shamika
shamika
16
Original post by aznkid66
Lol, way to trip right in front of the finish line..

EDIT: Right, ignore this then.


You could've been a little nicer about it, even when you thought he was wrong :tongue:

Do you see why und was right though?

Spoiler

Original post by Lord of the Flies
...

Original post by Star-girl
...


Both too quick ;_;
Original post by und
Problem 32

Find all positive integers nn such that 12n−11912n-119 and 75n−53975n-539 are both perfect squares.


By any chance is your next question going to be:

Spoiler

(edited 11 years ago)
Reply 146
Avatar for und
und
OP
16
Original post by metaltron
By any chance is your next question going to be:

Spoiler


Nah, I wasn't planning to post that one. :tongue:
Problem 34*

Prove that for all primes p>3, 24∣(p2−1).p>3, \ 24|(p^2-1).
Solution 34:

p2−1=(p−1)(p+1)p^2-1 = (p-1)(p+1)

P is prime and greater than 3, so p−1p-1 and p+1p+1 must contain factors of 2, 3 and 4, so it is divisible by 24.
Original post by Star-girl
Problem 34*

Prove that for all primes p>3, 24∣(p2−1).p>3, \ 24|(p^2-1).


p2−1=(p+1)(p−1) p^2 - 1 = (p+1)(p-1)

Since p>3, p cannot be divisible by 2 or 3. Hence, both p-1 and p+1 are even and one must be divisible by 4. Also, one of the three numbers, not p obviously, must be divisible by 3. Hence the whole expression is divisible by:

2×4×3=24 2 \times 4 \times 3 = 24

Edit : Dammit, I typed this so quickly as well!!
Original post by metaltron
p2−1=(p+1)(p−1) p^2 - 1 = (p+1)(p-1)

Since p>3, p cannot be divisible by 2 or 3. Hence, both p-1 and p+1 are even and one must be divisible by 4. Also, one of the three numbers, not p obviously, must be divisible by 3. Hence the whole expression is divisible by:

2×4×3=24 2 \times 4 \times 3 = 24

Edit : Dammit, I typed this so quickly as well!!


:colone:

(Let me have this one, I haven't got to any of the other ones first yet! :lol:)
Original post by DJMayes
:colone:

(Let me have this one, I haven't got to any of the other ones first yet! :lol:)


Either have I :frown: Of course you can have it, you beat me to it!
Reply 152
Avatar for und
und
OP
16
I was just casually typing up the solution, but you guys... :rolleyes:
Original post by und
I was just casually typing up the solution, but you guys... :rolleyes:


... were not doing it casually.
^^ I still haven't got to any one of them first. :wink:

Also nicely done, DJ and metaltron. :biggrin:
Problem 35*

Evaluate ∫0π2(xsin⁡x)2dx\displaystyle \int_0^{\frac{\pi}{2}} \bigg(\frac{x}{\sin x}\bigg)^2dx
Problem 36: */**

A particle is projected from the top of a plane inclined at an angle ϕ \phi to the horizontal. It is projected down the plane. Prove that; if the particle is to attain it's maximum range, the angle of projection θ \theta from the horizontal must satisfy:

θ=π4−ϕ2 \theta = \dfrac{\pi}{4} - \dfrac{\phi}{2}
(edited 11 years ago)
Reply 157
Avatar for und
und
OP
16
Original post by DJMayes
Problem 36: */**

A particle is projected from the top of a plane inclined at an angle ϕ \phi to the horizontal. It is projected down the plane. Prove that; if the particle is to attain it's maximum range, the angle of projection θ \theta must satisfy:

θ=π4−ϕ2 \theta = \dfrac{\pi}{4} - \dfrac{\phi}{2}

Isn't this a STEP question you were discussing recently?
Original post by und
Isn't this a STEP question you were discussing recently?


No; there was one involving firing up an inclined plane but it wasn't this. I did go to put a question on this thread about proving the maximum range of a projectile occurred when the angle of projection was 45 degrees but figured that it was too easy so messed about a bit and came up with this.
Original post by Lord of the Flies
Problem 35*

Evaluate ∫0π2(xsin⁡x)2dx\displaystyle \int_0^{\frac{\pi}{2}} \bigg(\frac{x}{\sin x}\bigg)^2dx

Solution 35
I hope!

35 Solution

(edited 11 years ago)

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