Question

What is `f(x) = int x-cscx dx` if `f((5pi)/4) = 0`?

Answer 1

`f(x) = (x^2)/2-ln(abs(csc(x) - cot(x))) + ln(sqrt(2)+1)-(25pi^2)/32`

Explanation:

`intx -csc(x) dx`
`=int x dx - int csc(x) dx`
`=(x^2)/2-ln(abs(csc(x) - cot(x))) + C`

Where `C` is a constant of integration.

So `f(x) = (x^2)/2-ln(abs(csc(x) - cot(x))) + C`

Using the fact that `f((5pi)/4) = 0`, we can substitute:

`f((5pi)/4)=(((5pi)/4)^2)/2-ln(abs(csc((5pi)/4) - cot((5pi)/4))) + C`

Moreover,

`(((5pi)/4)^2)/2-ln(abs(csc((5pi)/4) - cot((5pi)/4))) + C = 0`

Simplifying, we get

`(25pi^2)/32-ln(abs(-sqrt(2)-1))+C=0`
`(25pi^2)/32-ln(sqrt(2)+1)+C=0`
`C=ln(sqrt(2)+1)-(25pi^2)/32`

Therefore, the answer is:
`f(x) = (x^2)/2-ln(abs(csc(x) - cot(x))) + ln(sqrt(2)+1)-(25pi^2)/32`

Here's an image of what the graph looks like: