Approximating the integral $\int_0^{0.1} \sqrt{1-1/2\sin^2(t)} dt$

Question

How would I go about evaluating $I = \int_0^{0.1} \sqrt{1-1/2\sin^2(t)} dt$ to four decimal accuracy?

Answer 1

You can get 4 decimal places by a simple mid-point approximation:

$$I \approx 0.1 \sqrt{1-0.5 \sin^2{(0.05)}} \approx 0.0999375$$

where a numerical integration gives $I \approx 0.0999168$.

Answer 2

$I = \int_0^{0.1} \sqrt{1-1/2\sin^2(t)} dt \leq \int_0^{0.1} (1-1/4\sin^2(t)) dt = \frac{7}{80}+\frac{\sin(0.2)}{16} = 0.09991683317469133..$. The first inequality is since $\sqrt{1-x} \leq 1-\frac{x}{2}$ in your range of integration.

Also, since $\sin(x) \leq x$ in your range of integration, $I \geq \int_0^{0.1} \sqrt{1-t^2/2} dt = \frac{x\sqrt{2-x^2}-2\sin^{-1}(x/\sqrt{2})}{2\sqrt{2}} |_{x=0.1} = 0.09991660405478747...$.

So, $.09991660405478747... \leq I \leq 0.09991683317469133..$

Of course this assumes you can evaluate sines and arcsines and stuff, but perhaps you can use Taylor series, etc. for them.