Question

If \( a = \int_{\pi/3}^{\pi/6} \dfrac{\sin t + \cos t}{\sqrt{\sin 2t}} \, dt, \) then the value of \( \left( 2 \sin \frac{\alpha}{2} + 1 \right)^2 \text{ is ............}. \)

Hint:

For integrals involving trigonometric functions, use trigonometric identities to simplify the expressions before integrating.

Answer 1

Correct Answer: 2.9 - 3.1

To find \( a = \int_{\pi/3}^{\pi/6} \dfrac{\sin t + \cos t}{\sqrt{\sin 2t}} \, dt \), we need to simplify the integrand. First, note \( \sin 2t = 2 \sin t \cos t \), so \( \sqrt{\sin 2t} = \sqrt{2 \sin t \cos t} = \sqrt{2} \sqrt{\sin t} \sqrt{\cos t} \).

Thus, the integrand becomes \(\dfrac{\sin t + \cos t}{\sqrt{2} \sqrt{\sin t \cos t}}\). We can simplify further with a substitution. Let \( u = \frac{\pi}{2} - t \), yielding \( du = -dt \). This changes the integral bounds: when \( t = \pi/3 \), \( u = \pi/6 \), and when \( t = \pi/6 \), \( u = \pi/3 \).

The integral \( a \) becomes:

\( a = \int_{\pi/6}^{\pi/3} \dfrac{\cos u + \sin u}{\sqrt{2} \sqrt{\cos u \sin u}} \, (-du) = \int_{\pi/6}^{\pi/3} \dfrac{\cos u + \sin u}{\sqrt{2} \sqrt{\sin 2u/2}} \, (-du) = \int_{\pi/6}^{\pi/3} \dfrac{\cos u + \sin u}{\sin u + \cos u} \, du \)

This simplifies to \( a = \int_{\pi/6}^{\pi/3} du = \left[ u \right]_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \).

We now need to evaluate \( \left( 2 \sin \frac{\alpha}{2} + 1 \right)^2 \) with \( \alpha = \frac{\pi}{6} \). We find:

\(\sin \frac{\alpha}{2} = \sin \frac{\pi}{12} = \frac{\sqrt{6} - \sqrt{2}}{4}\), derived using sin addition formulas.

Thus, \( 2 \sin \frac{\pi}{12} = \frac{\sqrt{6} - \sqrt{2}}{2} \), and:

\(2 \sin \frac{\pi}{12} + 1 = \frac{\sqrt{6} - \sqrt{2} + 2}{2} = \frac{\sqrt{6} - \sqrt{2} + 2}{2}\).

Square it to find:

\(\left(\frac{\sqrt{6} - \sqrt{2} + 2}{2}\right)^2 = \frac{(\sqrt{6} + 2 - \sqrt{2})^2}{4} = \frac{(\sqrt{6})^2 + 2(\sqrt{6})(2) - 2(\sqrt{6})(\sqrt{2}) + 4 + (\sqrt{2})^2}{4} = 3 + 2\sqrt{12} - 2\sqrt{12} + 4 + 2 = \frac{9}{4}\) which is approximately \(3.0625\).

Answer: 3.0625