Arc Length Calculation-Not sure how to calculate intergal

[ku1005]

Hi, just attempting this arc length calculation, but not sure how to calculate the intergal at the end, because usually, the square root will disappear (collapse down due to manipulation of equation therefore making it simple to calculate the intergal).In this case, I have either doen something wrong or am not sure how to, any tips on my working would be greatly appreciated! cheers

Arc Length =the intergal from a to b of the sqrt(1+[f'(x)]^2)

So for



i have



Thus
Arc Length



However im not sure how to calculate this, and as I said, usually it collapses down to something simpler...

i was thinking along the lines of



but that still doesnt help really....so yeah...if anyone has any tips on my workings cheers!

 

[morson]

= \(\displaystyle \L\ \int_0^{1} \sqrt{ \frac{4 - x^2}{4 - x^2}\ + \frac{x^2}{4 - x^2}\ } dx\)

= \(\displaystyle \L\ \int_0^{1} \sqrt{ \frac{4}{4 - x^2}\ } dx \ = \int_0^{1} \frac{2 dx}{\sqrt{4 - x^2}\\\)

Now let x = 2u, so dx = 2du

When x = 1, u = 0.5. When x = 0, u = 0.

= \(\displaystyle \L\ \int_0^{\frac{1}{2}} \frac{2du}{\sqrt{1 - u^2}\\\)

= \(\displaystyle \L\ 2\int_0^{\frac{1}{2}} \frac{du}{\sqrt{1 - u^2}\\\)

What's the derivative of arcSin(u)? Remember, that we're dealing with radians, too.

 

[ku1005]

cheers....rest is easy..it was just the step where you collapsed/got rid of the square root sign I had the problem with.thanks for your help

 

[Deleted member 4993]

Whenever you see "square-root" and "x^2" - try trigonometric substitution. For example:

If you have:

sqrt(a^2 - x^2) ....try x = a* sin(u)

sqrt(a^2 + x^2) ....try x = a* tan(u)