Understanding the Inequality in Trigonometric Function Analysis

gajohnson
Messages
70
Reaction score
0

Homework Statement



This isn't really a homework question, just working through Rudin and got caught up on something. C(x) and S(x) refer to cos(x) and sin(x) respectively.

Here is the section in question:

http://grab.by/mSo8

Homework Equations


The Attempt at a Solution



Well the part I'm having trouble understanding is the claim: "Hence, if 0≤x≤y, we have S(x)(y-x)<\int^{y}_{x}{S(t)}dt = C(x)-C(y)≤2"

In particular, the inequality S(x)(y-x)<\int^{y}_{x}{S(t)}dt is not clear to me. I reviewed a number of integration theorems but couldn't come up with anything that states this. Any help understanding how this inequality is derived would be much appreciated!

EDIT: OK, this might be really obvious. Is this simply true by the definition of the Riemann integral?
 
Last edited:
Physics news on Phys.org
It follows from the observation that S(t) is strictly increasing on the interval, so within the interval S(x) < S(t). Then integrate both sides over the interval.
 
  • Like
Likes 1 person
Ah, of course. Got it. Thanks!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top