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Integration Maths Problem

Can anyone help me solve:
Int (sqrt(x+1)-sqrt(x))^(3/2) dx.
With substitution, it becomes of the form:

Int (cosh(y)-sinh(y))^(3/2)/sinh(2y) dy
or 1//2 Int exp(-3y/2)/(exp(2y)-exp(-2y)) dy

Even if you cannot integrate this function, ideas would be extremely helpful.
(edited 12 years ago)
Reply 1
i managed to transform the last version into

int 1/(w8 - 1) dw

but am not sure what to do after that
Might I ask how you got that?
Wolfram Alpha cannot do this (at least in the amount of time I am entitled to on their servers!)
Reply 4
Original post by Daaaaaaaaaaaaaaaan
Might I ask how you got that?


i cannot reveal all the details due to TSR policy... but basically you let

w = ey/2
Reply 5
Original post by Daaaaaaaaaaaaaaaan
Can anyone help me solve:
Int (sqrt(x+1)-sqrt(x))^(3/2) dx.
With substitution, it becomes of the form:

Int (cosh(y)-sinh(y))^(3/2)/sinh(2y) dy
or 1//2 Int exp(-3y/2)/(exp(2y)-exp(-2y)) dy

Even if you cannot integrate this function, ideas would be extremely helpful.


I thought this forum had LaTeX support
Reply 6
Original post by rhysowen
I thought this forum had LaTeX support


LaTex support sounds like something Granny would purchase from Boots
Reply 8
Original post by the bear
i managed to transform the last version into

int 1/(w8 - 1) dw

but am not sure what to do after that
Assuming this is correct, you can write

w^8 - 1 = (w^4+1)(w^4-1) = (w^4+1)(w^2+1)(w^2-1).

Then w4+1=(w2w2+1)(w2+w2+1)w^4+1 = (w^2-w\sqrt{2} +1)(w^2+w\sqrt{2}+1).

Then some (painful) partial fractioning will finish things off.
Reply 9
Original post by Daaaaaaaaaaaaaaaan
Can anyone help me solve:
Int (sqrt(x+1)-sqrt(x))^(3/2) dx.
With substitution, it becomes of the form:

Int (cosh(y)-sinh(y))^(3/2)/sinh(2y) dy
or 1//2 Int exp(-3y/2)/(exp(2y)-exp(-2y)) dy

Even if you cannot integrate this function, ideas would be extremely helpful.


For (x+1x)32dx \displaystyle \int (\sqrt{x+1}- \sqrt{x})^{\frac{3}{2}}dx
Use the following substitution

x=sinh2tx=sinh^2 t
x+1=1+sinh2t=cosh2tx+1=1+sinh^2 t=cosh^2 t
dx=2sinhtcosht=sinh2tdx=2sinht \cdot cosht =sinh 2t
So
(coshtsinht)322sinh2tdt\displaystyle \int (cosh t - sinh t)^{\frac{3}{2}} 2sinh2t dt

using
cosht=et+et2\displaystyle cosht =\frac{e^t+e^{-t}}{2}
sinht=etet2\displaystyle sinht =\frac{e^t-e^{-t}}{2}

the
(coshtsinht)32=(12(et+etet+et))32=e32t\displaystyle (cosh t-sinht)^{\frac{3}{2}}=\left (\frac{1}{2} \left (e^t+e^{-t}-e^t+e^{-t}\right )\right )^{\frac{3}{2}}=e^{-\frac{3}{2}t}

So
12e32t(e2te2t)dt=\displaystyle \frac{1}{2} \int e^{-\frac{3}{2}t} \cdot \left (e^{2t}-e^{-2t}\right ) dt =
12(et2e72t)dt\displaystyle \frac{1}{2} \int \left (e^\frac{t}{2} - e^{-\frac{7}{2}t}\right ) dt

Now You can integrate this
(edited 12 years ago)
Reply 10
Well done ztibor & DFranklin :wink:
Thank you :smile:

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