Q. The value of $\alpha$ which satisfying $\int_{\frac{\pi}{2}}^{\alpha}{\sin{x\ dx = \sin{2\alpha}}},\ \alpha \in (0,\ 2\pi)$ is equal to
A) $\frac{\pi}{2}$
B) $\frac{3\pi}{2}$
C) $\frac{7\pi}{6}$
D) All of these
Explanation
(d)
$\int_{\frac{\pi}{2}}^{\alpha}{\sin{x\ dx = \sin{2\alpha}}},\ \alpha \in (0,\ 2\pi)$
$\Rightarrow \ - \left\lbrack \cos x \right\rbrack_{\frac{\pi}{2}}^{\alpha} = \sin{2\alpha}$
$\Rightarrow \ - \cos\alpha = \sin{2\alpha}$
$\Rightarrow$ $\sin\left( \alpha - \frac{\pi}{2} \right) = \sin{2\alpha}$
Hence, $\alpha = \frac{\pi}{2},\frac{3\pi}{2},\frac{7\pi}{6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.