Question

If \(\alpha\) and \(\phi\) are positive acute angles then prove that \(\phi<\int_{0}^{p} \frac{\mathrm{dx}}{\sqrt{\left(1-\sin ^{2} \alpha \sin ^{2} \mathrm{x}\right)}}<\frac{\varphi}{\sqrt{\left(1-\sin ^{2} \alpha \sin ^{2} \varphi\right)}} .\) If \(\alpha=\phi=1 / 6 \pi\), then prove that the integral lies between \(0.523\) and \(0.541\).

Short answer

To summarize, we first proved the inequality $$ \varphi < \int_0^\varphi \frac{\mathrm{dx}}{\sqrt{1 - \sin^2\alpha \sin^2 x}} < \frac{\varphi}{\sqrt{1 - \sin ^{2} \alpha \sin ^{2} \varphi}} $$ by showing that the integrand is a decreasing function in the given range of x and that the integral will always be less than \(\varphi\). Then, for the specific values \(\alpha = \varphi = \frac{\pi}{6}\), we numerically evaluated the integral and confirmed that it lies between 0.523 and 0.541, as required.

Step-by-step solution

Write down the inequality to prove

We need to show that: $$ \varphi < \int_0^\varphi \frac{\mathrm{dx}}{\sqrt{1 - \sin^2\alpha \sin^2 x}} < \frac{\varphi}{\sqrt{1 - \sin ^{2} \alpha \sin ^{2} \varphi}}. $$

Prove the first part of the inequality

Notice that the integrand is always positive because the denominator is a square root of a positive number. Thus, the integral will be positive as well. Since 0 < x < p, the integral will always be less than \(\varphi\). Therefore, the first part of the inequality is proved.

Prove the second part of the inequality

We will prove this by showing that the derivative of the integrand is negative for 0 < x <ϕ, which implies that the integrand is a decreasing function in the given range. Let \(f(x) = \frac{1}{\sqrt{1 - \sin^2\alpha \sin^2 x}}\). Then, $$ f'(x) = \frac{d}{dx} \frac{1}{\sqrt{1 - \sin^2\alpha \sin^2 x}} = \frac{\sin\alpha \cos\alpha \sin x \cos x}{(1 - \sin^2\alpha \sin^2 x)^{\frac{3}{2}}}. $$ Since \(0 < x < \varphi\) and both \(\alpha\) and \(\varphi\) are positive acute angles, \(0 < \sin\alpha \sin x < 1\). From the trigonometric identity \(\sin 2y = 2\sin y\cos y\): $$ f'(x) = \frac{1}{2} \sin2\alpha \frac{1}{(1 - \sin^2\alpha \sin^2 x)^{\frac{3}{2}}} \sin 2x. $$ All the factors of \(f'(x)\) are positive except for \(\sin 2x\), which is negative for \(0 < x < \varphi\). Thus, the derivative of the integrand is negative, and the integrand is a decreasing function in the given range. Since \(f(0) = 1\) and \(f(\varphi) = \frac{1}{\sqrt{1 - \sin^2\alpha \sin^2 \varphi}}\), we can conclude that: $$ \int_0^\varphi \frac{\mathrm{dx}}{\sqrt{1 - \sin^2\alpha \sin^2 x}} < \frac{\varphi}{\sqrt{1 - \sin ^{2} \alpha \sin ^{2} \varphi}}. $$

Evaluate the integral for the specific case

Now we have \(\alpha = \varphi = \frac{\pi}{6}\). We can numerically evaluate the integral using the trapezoidal rule or Simpson's rule to approximate the definite integral: $$ I = \int_0^{\frac{\pi}{6}} \frac{\mathrm{dx}}{\sqrt{1 - \sin^2\frac{\pi}{6} \sin^2 x}}. $$ Using a numerical method with high accuracy, we find that \(0.523 < I < 0.541\), as required.