Question

If $\frac{\pi }{2}<t<\frac{2\pi }{3}$ and $={\int }_0^{\sin 2t}\frac{dx}{\sqrt{4{\cos }^{2}t-x^2}}$ then the value of $\frac{2005(I+t)}{\pi }$ equals

• A. $2004$
• B. $2006$
• C. $2005$
• D. None of these

Answer 1

The correct option is C $2005$

$I={\int }_0^{\sin 2t}\frac{dx}{\sqrt{{\left(2\cos t\right)}^2-x^2}}={\left\{{\sin }^{-1}\left(\frac{x}{2\cos t}\right)\right\}}_0^{\sin 2t}$
$={\sin }^{-1}\left(\frac{\sin 2t}{2\cos t}\right)={\sin }^{-1}\left(\sin t\right)={\sin }^{-1}\left(\sin (\pi -t)\right)$
$(\because \frac{\pi }{2}<t<\frac{2\pi }{3})$
$I=\pi -t$ $I+t=\pi (\because \frac{\pi }{3}<\pi -t<\frac{\pi }{2})$
$\therefore \frac{I+t}{\pi }=1$
$\therefore \frac{2005(I+t)}{\pi }=2005$