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I encountered the following integral question:

Show that

$$\frac{3}{8} \leq \int_{0}^{\frac{1}{2}} \frac{\sqrt{1-x}}{\sqrt{1+x}} \, dx \leq \frac{\sqrt{3}}{4}$$

I know I can just integrate it, but is there a way to obtain the result above without directly integrating? I tried doing rectangles and trapezoids but neither yield the result above.

Also do you have any tips for tackling with other such questions of a similar vein?

Kevin
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    For the lower bound, write your integrand as $\frac{\sqrt{1-x^2}}{1+x}$ and $\sqrt{1-x^2} \ge 1-x^2$ when $0\le x \le 1/2$. Thus, your integral is bounded below by $$\int_0^{1/2}\frac{1-x^2}{1+x}dx=\int_0^{1/2}(1-x)dx = 3/8.$$ Similarly, your integral is no more than $\int_0^{1/2}\frac{1}{1+x}dx = \log(3/2)\approx 0.4055$. This gives you a better upper bound then $\sqrt{3}/4 \approx 0.433$. – Quoka Mar 14 '24 at 23:08
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    @Quoka thx! I get how to do it now. Do you want to write an answer so I can accept it? – Kevin Mar 15 '24 at 03:54

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I'll transcribe my comment on the lower bound here. Observe that the integrand is precisely $$ \frac{\sqrt{1-x^2}}{1+x}. $$ Since $\sqrt{1-x^2} \ge 1-x^2$ on the interval $[0,1/2]$, we have $$ \int_0^{1/2}\frac{\sqrt{1-x^2}}{1+x}dx \ge \int_0^{1/2}\frac{1-x^2}{1+x}dx=\int_0^{1/2}(1-x)dx = 3/8. $$

Quoka
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