Question

What is the arclength of `(tant-sect*csct)` on `t in [pi/8,pi/3]`?

Answer 1

`L=cot(pi/8)-cot(pi/3)+4sum_(n=1)^oo((1/2),(n))1/16^n{((4n-2),(2n-1))(5pi)/24+sum_(k=0)^(2n-2)((4n-2),(k))(-1)^(2n-1-k)/(2n-1-k)(sin((4n-2-2k)pi/3)-sin((4n-2-2k)pi/3))}` units.

Explanation:

`f(t)=tant-sect*csct=-cott`

`f'(t)=csc^2t`

Arclength is given by:

`L=int_(pi/8)^(pi/3)sqrt(1+csc^4t)dt`

Rearrange:

`L=int_(pi/8)^(pi/3)csc^2tsqrt(1+sin^4t)dt`

Since `sin^4t<1`, take the series expansion of the square root:

`L=int_(pi/8)^(pi/3)csc^2t{sum_(n=0)^oo((1/2),(n))sin^(4n)t}dt`

Isolate the `n=0` term:

`L=int_(pi/8)^(pi/3)csc^2tdt+sum_(n=1)^oo((1/2),(n))int_(pi/8)^(pi/3)sin^(4n-2)tdt`

Apply the trigonometric power-reduction formula:

`L=[-cott]_ (pi/8)^(pi/3)+sum_(n=1)^oo((1/2),(n))int_(pi/8)^(pi/3){1/2^(4n-2)((4n-2),(2n-1))+2/2^(4n-2)sum_(k=0)^(2n-2)((4n-2),(k))(-1)^(2n-1-k)cos((4n-2-2k)t)}dt`

Simplify:

`L=cot(pi/8)-cot(pi/3)+4sum_(n=1)^oo((1/2),(n))1/16^nint_(pi/8)^(pi/3){((4n-2),(2n-1))+2sum_(k=0)^(2n-2)((4n-2),(k))(-1)^(2n-1-k)cos((4n-2-2k)t)}dt`

Integrate directly:

`L=cot(pi/8)-cot(pi/3)+4sum_(n=1)^oo((1/2),(n))1/16^n[((4n-2),(2n-1))t+sum_(k=0)^(2n-2)((4n-2),(k))(-1)^(2n-1-k)/(2n-1-k)sin((4n-2-2k)t)]_(pi/8)^(pi/3)`

Simplify:

`L=cot(pi/8)-cot(pi/3)+4sum_(n=1)^oo((1/2),(n))1/16^n{((4n-2),(2n-1))(5pi)/24+sum_(k=0)^(2n-2)((4n-2),(k))(-1)^(2n-1-k)/(2n-1-k)(sin((4n-2-2k)pi/3)-sin((4n-2-2k)pi/3))}`