How Can the Inequality -x ≤ sin(x) ≤ x Help Prove a Convergent Integral?

[henry22]

Homework Statement

I am attempting to show that $-x \leq sin(x) \leq x$ for x>0 and thus $\int^1_0 nxsin(\frac{1}{nx})dx$ converges to 1.

Homework Equations

I know that I need to use the fact that I have shown that the limit as T tends to infinity of $\int^T_1 \frac{cos(x)}{\sqrt{x}}dx$ exists.

The Attempt at a Solution

 

[LCKurtz]

Homework Statement

I am attempting to show that $-x \leq sin(x) \leq x$ for x>0 and ...

The -x part is trivial and for the rest integrate both sides of cos(t) ≤ 1 between 0 and x, x > 0.

 

[henry22]

The -x part is trivial and for the rest integrate both sides of cos(t) ≤ 1 between 0 and x, x > 0.

OK I've done this and I get the inequality I need. But can I just check, I don't understand how I have used the equation I need to in the OP?

For the second part if I know that -x<= sinx <= x then -1<=nx(sin(1/nx)) <= 1 but then I'm a bit stuck