intx -csc(x) dx∫x−csc(x)dx
=int x dx - int csc(x) dx=∫xdx−∫csc(x)dx
=(x^2)/2-ln(abs(csc(x) - cot(x))) + C=x22−ln(|csc(x)−cot(x)|)+C
Where CC is a constant of integration.
So f(x) = (x^2)/2-ln(abs(csc(x) - cot(x))) + Cf(x)=x22−ln(|csc(x)−cot(x)|)+C
Using the fact that f((5pi)/4) = 0 f(5π4)=0, we can substitute:
f((5pi)/4)=(((5pi)/4)^2)/2-ln(abs(csc((5pi)/4) - cot((5pi)/4))) + Cf(5π4)=(5π4)22−ln(∣∣∣csc(5π4)−cot(5π4)∣∣∣)+C
Moreover,
(((5pi)/4)^2)/2-ln(abs(csc((5pi)/4) - cot((5pi)/4))) + C = 0(5π4)22−ln(∣∣∣csc(5π4)−cot(5π4)∣∣∣)+C=0
Simplifying, we get
(25pi^2)/32-ln(abs(-sqrt(2)-1))+C=025π232−ln(∣∣−√2−1∣∣)+C=0
(25pi^2)/32-ln(sqrt(2)+1)+C=025π232−ln(√2+1)+C=0
C=ln(sqrt(2)+1)-(25pi^2)/32C=ln(√2+1)−25π232
Therefore, the answer is:
f(x) = (x^2)/2-ln(abs(csc(x) - cot(x))) + ln(sqrt(2)+1)-(25pi^2)/32f(x)=x22−ln(|csc(x)−cot(x)|)+ln(√2+1)−25π232
Here's an image of what the graph looks like:
useruploads.socratic.org)
