Question

What is the net area between `f(x) = cotxcscx -sinx` and the x-axis over `x in [pi/6, (5pi)/8 ]`?

Answer 1

`A=A_1+A_2=2-1/sin((5pi)/8)+cos((5pi)/8)-sqrt(3)/2~~-.331`

Explanation:

`A = int_(pi/6)^((5pi)/8) (cotxcscx - sinx) dx`
`A_1+A_2=int_(pi/6)^((5pi)/8) cotxcscx dx - int_(pi/6)^((5pi)/8)sinx dx`
`A_1 = int_(pi/6)^((5pi)/8) cotxcscx dx`
`int_(pi/6)^((5pi)/8) cosx/sinx*1/sinx dx= int_(pi/6)^((5pi)/8) cosx/sin^2x dx`

Let `u = sinx => (du)/dx= cosx` then rewrite the integral as:
`A_1=int (du)/u^2= -1/u` substitute u=sinx
`A_1=-1/sinx= -[cscx]_(pi/6)^((5pi)/8) = 2-1/sin((5pi)/8)`

`A_2=int_(pi/6)^((5pi)/8)sinx dx= [-cosx]_(pi/6)^((5pi)/8)`
`A_2= cos((5pi)/8) -sqrt(3)/2`

`A=A_1+A_2=2-1/sin((5pi)/8)+cos((5pi)/8)-sqrt(3)/2~~-.331`