Q. $\int_{}^{}{\sin\sqrt{3}}\ dx$ is equal to
A) $\sin\sqrt{x} - \sqrt{x}\cos\sqrt{x}$
B) $2\left( \sin\sqrt{x} - \sqrt{x}\cos\sqrt{x} \right) + c$
C) $\cos\sqrt{x} - \sqrt{x}\sin\sqrt{x} + c$
D) $2\left( \cos\sqrt{x} - \sqrt{x}\sin\sqrt{x} \right) + c$
Explanation
(b)
Let $I = \int_{}^{}{\sin\sqrt{3}}\ dx$
Put $\sqrt{x} = t\ \Rightarrow \ \frac{1}{2\sqrt{x}}dx = dt$
$\therefore\ \ I = \int_{}^{}{2t\sin{t\ dt}}$
$= 2\lbrack - t\cos t + \int_{}^{}{\cos{t\ dt}}\rbrack$
$= 2\left\lbrack - t\cos t + \sin t \right\rbrack + c$
$= 2\left\lbrack - \sqrt{x}\cos\sqrt{x} + \sin\sqrt{x} \right\rbrack + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.