MacLaurin Approximation for 1-cos(x)/x

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integrate from 0 -> 1
(1-cosx)/x dx

find maclorein approximation
this has been killing me any ideas??
 
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Start with the Maclauren series for cosx, do you know it?
 
rock.freak667 said:
Start with the Maclauren series for cosx, do you know it?

well one of my issues atm that forgot to add on there is the first term is over zero isn't it? due to the lower bounds and the serious should be 1-1/2x^2 +1/24x^4...
if that helps you find where i am lost :)
 
so

cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...

what is 1-cos(x)? Then what is (1-cosx)/x ?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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