MCQ
$\int_{}^{} {\sqrt {2 + \sin 3x} \;.\;\cos 3x\;dx = } $
  • A
    $\frac{2}{9}{(2 + \sin 3x)^{1/2}} + c$
  • B
    $\frac{2}{3}{(2 + \sin 3x)^{2/3}} + c$
  • C
    $\frac{2}{3}{(2 + \sin 3x)^{3/2}} + c$
  • $\frac{2}{9}{(2 + \sin 3x)^{3/2}} + c$

Answer

Correct option: D.
$\frac{2}{9}{(2 + \sin 3x)^{3/2}} + c$
d
(d) Put $2 + \sin 3x = t \Rightarrow 3\cos 3x\,dx = dt,$ then
$\int_{}^{} {\sqrt {2 + \sin 3x} \cos 3x\,dx} = \frac{1}{3}\int_{}^{} {{t^{1/2}}dt} = \frac{2}{9}{t^{3/2}} + c$
$ = \frac{2}{9}{(2 + \sin 3x)^{3/2}} + c$.

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The correct option is:

Let $A = \left[ {\begin{array}{*{20}{c}}
  2&b&1 \\ 
  b&{{b^2} + 1}&b \\ 
  1&b&2 
\end{array}} \right]$  where $b > 0$. Then the minimum value of $\frac{{\det \left( A \right)}}{b}$ is