Question

The value of $\int\limits_{\pi/6}^{5\pi/6}\sqrt{4-4\,\sin^2\,t}\,dt$ is

• A. 0
• B. 2
• C. 1
• D. none of these.

Answer 1

Answer: (B)

$I=\int\limits_{\pi 6}^{5 \pi /6}\sqrt{4-4 sin^{2}t}dt=2 \int\limits_{\pi/ 6}^{5\pi /6}\left|cos\,t\right|dt$
$=2 \int\limits_{\pi /6}^{\pi /2}\left|cos\,t\right|dt+2\int\limits^{5\pi/ 6}_{\pi /2}\left|cos\,t\right|dt$
$=2 \int\limits_{\pi/ 6}^{\pi /2} cos\,t \, dt+2 \int\limits_{\pi/ 2}^{5 \pi /6}-cos\, t \, dt$
$=2\cdot\left|sin\,t\right|_{\pi /6}^{\pi/ 2}-2 \left|sin\,t\right|_{\pi /2}^{5\pi/ 6}$
$=2 \left[sin \frac{\pi}{2}-sin \frac{\pi}{6}\right]-\left[sin \frac{5\pi}{6}-sin \frac{\pi}{2}\right]$
$=2\left(1-\frac{1}{2}\right)-2\left(\frac{1}{2}-1\right)$
$=2\left(\frac{1}{2}\right)-2\left(-\frac{1}{2}\right)=1+1=2$