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

1 Answer
Jul 16, 2018

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:
WolframAlpha.comWolframAlpha.com