Define the function
Establish the following:
- .
- if .
- if .
- Proof. We consider ,
This is then greater than 0 since for (hence, the integrand is positive everywhere on the interval of integration, so the integral is positive). Thus,
- Proof. We compute, recalling the trig identity and that the derivative of is ,
Now, we make the substitution , and so,
- Proof. From part (a) we have so ; hence,
From part (b) we know . Therefore,
For the inequality on the right we again use the inequality in part (a), and so,
Using part (b) again we know . Hence,
Putting these together we have
Comparison theorem, that I think you are using here, has less-than-or-equal relationship, and not strictly less-than. Where are you basing your assumption on?