Question

What is the arc length of `f(x) = sinx` on `x in [pi/12,(5pi)/12]`?

Answer 1

`L=pi/3+sum_(n=1)^oo((1/2),(n))1/4^n{((2n),(n))pi/3+2sum_(k=0)^(n-1)((n),(k))(sin((n-k)pi/3)cos((n-k)pi/2))/(n-k)}`

Explanation:

`f(x)=sinx`

`f'(x)=cosx`

Arc length is given by:

`L=int_(pi/12)^((5pi)/12)sqrt(1+cos^2x)dx`

For `x in [pi/12,(5pi)/12]`, `cos^2x<1`. Take the series expansion of the square root:

`L=int_(pi/12)^((5pi)/12)sum_(n=0)^oo((1/2),(n))cos^(2n)xdx`

Isolate the `n=0` case:

`L=int_(pi/12)^((5pi)/12)dx+sum_(n=1)^oo((1/2),(n))int_(pi/12)^((5pi)/12)cos^(2n)xdx`

Apply the Trigonometric power-reduction formula:

`L=[x]_ (pi/12)^((5pi)/12)+sum_(n=1)^oo((1/2),(n))int_(pi/12)^((5pi)/12){1/4^n((2n),(n))+2/4^nsum_(k=0)^(n-1)((n),(k))cos(2(n-k)x)}dx`

Simplify:

`L=pi/3+sum_(n=1)^oo((1/2),(n))1/4^nint_(pi/12)^((5pi)/12){((2n),(n))+2sum_(k=0)^(n-1)((n),(k))cos(2(n-k)x)}dx`

Integrate directly:

`L=pi/3+sum_(n=1)^oo((1/2),(n))1/4^n[((2n),(n))x+sum_(k=0)^(n-1)((n),(k))sin(2(n-k)x)/(n-k)]_(pi/12)^((5pi)/12)`

Simplify:

`L=pi/3+sum_(n=1)^oo((1/2),(n))1/4^n{((2n),(n))pi/3+sum_(k=0)^(n-1)((n),(k))(sin((n-k)(5pi)/6)-sin((n-k)pi/6))/(n-k)}`

Apply the Trigonometric sum-to-product identity `sina-sinb=2sin((a-b)/2)cos((a+b)/2)`:

`L=pi/3+sum_(n=1)^oo((1/2),(n))1/4^n{((2n),(n))pi/3+2sum_(k=0)^(n-1)((n),(k))(sin((n-k)pi/3)cos((n-k)pi/2))/(n-k)}`

Isolate the `n=1` case:

`L=pi/3+pi/12+sum_(n=2)^oo((1/2),(n))1/4^n{((2n),(n))pi/3+2sum_(k=0)^(n-1)((n),(k))(sin((n-k)pi/3)cos((n-k)pi/2))/(n-k)}`