MCQ
$\int_0^2 {\sqrt {\frac{{2 + x}}{{2 - x}}} } \,dx = $
- ✓$\pi + 2$
- B$\pi + \frac{3}{2}$
- C$\pi + 1$
- DNone of these
$\Rightarrow dx = - 2\sin \theta \,d\theta ,$ then
$\int_0^2 {\sqrt {\frac{{2 + x}}{{2 - x}}} } dx = - 2\int_{\pi /2}^0 {\sqrt {\frac{{1 + \cos \theta }}{{1 - \cos \theta }}} } \sin \theta \,d\theta $
$ = 4\int_0^{\pi /2} {\frac{{\cos (\theta /2)}}{{\sin (\theta /2)}}\sin \frac{\theta }{2}\cos \frac{\theta }{2}d\theta } $
$ = 2\int_0^{\pi /2} {(1 + \cos \theta )\,d\theta } $
$ = 2[\theta + \sin \theta ]_0^{\pi /2} $
$= 2\left[ {\frac{\pi }{2} + 1} \right] = \pi + 2$.
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$f (\theta)=\left|\begin{array}{ccc}-\sin ^{2} \theta & -1-\sin ^{2} \theta & 1 \\ -\cos ^{2} \theta & -1-\cos ^{2} \theta & 1 \\ 12 & 10 & -2\end{array}\right|$ are $m$ and $M$ respectively, then the ordered pair $( m , M )$ is equal to