Prove Integral \int_{0}^{x}\frac{\sin(t)}{t+1}dt > 0 Without Calculus

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Homework Statement



Prove that \int_{0}^{x}\frac{\sin(t)}{t+1}dt > 0

Homework Equations



It asks me to do this without actually calculating it and I can't use the Fundamental Theorem of Calculus.

The Attempt at a Solution



Okay, I see why this should be true as the function oscillates (with smaller bumps over time) around the x-axis so only the first bump really counts. I've tried considering small intervals where I know the function has certain values, but I can't seem to get anywhere with that idea either. Any help?
 
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Even a hint would be greatly appreciated! Thanks!
 
Ughhh I figured it out! I just showed that the integral from [0,pi] is larger in absolute value than the integral from [pi,2pi]. Continuing in this fashion (i.e. pairing up consecutive intervals) proves the assertion.
 
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...
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