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# P5 help! watch

1. 2 more questions to put to you lot!!

review exercise q49
show that the curve with the equation y=sechx has 2nd differental =0
at points where x= +/- lnp and state value of p

the next one... Q50

I[n]=integral of sin^n x dx between pi/2 and 0
a) show that (n+2)I[n+2]=(n+1)I[n] this bit was fine

show that I[2n]=(2n)!pi / 2^(2n+1)(n!)^2 which i really can't do!

and show further that I[2n-1]I[2n]=pi/4n

sorry people!!! thankyou
2. (Original post by ogs)
2 more questions to put to you lot!!

review exercise q49
show that the curve with the equation y=sechx has 2nd differental =0
at points where x= +/- lnp and state value of p

the next one... Q50

I[n]=integral of sin^n x dx between pi/2 and 0
a) show that (n+2)I[n+2]=(n+1)I[n] this bit was fine

show that I[2n]=(2n)!pi / 2^(2n+1)(n!)^2 which i really can't do!

and show further that I[2n-1]I[2n]=pi/4n

sorry people!!! thankyou
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3. y = sech(x)
dy/dx = -tanh(x)sech(x)
d^2y/dx^2
= -sech^3(x) + tanh^2(x)sech(x)
= sech^3(x) (-1 + sinh^2(x))

When d^2y/dx^2 = 0 we have sinh(x) = 1 or -1. If sinh(x) = 1 then

e^x - e^(-x) = 2
e^2x - 2e^x - 1 = 0
e^x = [2 + sqrt(4 + 4)] / 2 = 1 + sqrt(2)
x = ln(1 + sqrt(2)).

Similarly (or using the fact that sinh is odd) if sinh(x) = -1 then x = -ln(1 + sqrt(2))
4. For your second question see here: http://www.nrich.maths.org/discus/me...tml?1081469967.
5. I(2n)
= [(2n - 1)/2n] * I(2n - 2)
= [(2n - 1)/2n] * [(2n - 3)/(2n - 2)] * I(2n - 4)
= [(2n - 1)/2n] * [(2n - 3)/(2n - 2)] * [(2n - 5)/(2n - 4)] * I(2n - 6)
= ...
= [(2n - 1)(2n - 3) ... 1] I(0) / [2n (2n-2) ... 2]
= [(2n - 1)(2n - 3) ... 1] pi / {2[2n (2n-2) ... 2]}
= (2n)! pi / {2[2n (2n-2) ... 2]^2}
= (2n)! pi / {2^(2n + 1) (n!)^2}
6. (Original post by mikesgt2)
For your second question see here: http://www.nrich.maths.org/discus/me...tml?1081469967.
hah i remeber this!
7. Thankyou every one!

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