The one with the log and trigonometry.

Let the given integral be denoted as,

$\displaystyle I(k)=\int _{ 0 }^{ \pi /2 }{ \ln { { (sin }^{ 2 }(x)+{ k }^{ 2 } } { cos }^{ 2 }(x))dx }$

Now differentiating wrt $k$ ,

$\displaystyle \frac { d(I(k)) }{ d(k) } =\int _{ 0 }^{ \pi /2 }{ \frac { 2k{cos}^{2}x }{ { sin }^{ 2 }x+{ k }^{ 2 }{ cos }^{ 2 }x } } dx$

Now integrating the RHS wrt $x$ by writing $\displaystyle sin(x)=\frac { 2tan(x/2) }{ 1+{ tan }^{ 2 }(x/2) }$ and $\displaystyle cos(x)=\frac { 1-{ tan }^{ 2 }(x/2) }{ 1+{ tan }^{ 2 }(x/2) }$ .

Then,by taking $t=tan(x/2)$ and solving we get that,

$\displaystyle \frac { d(I(k)) }{ d(k) }=\dfrac{\pi}{(k+1)}$

Then,by integrating with wrt to $k$ we get that,

$I(k)=\pi*ln|k+1|+c$

Now,by substituting $k=1$ in original integral we get that,

$I(1)=0=\pi*ln(2)+c$ which gives $c=-\pi*ln(2)$

Therefore the integral is,

$I(k)=\pi*ln|k+1|-\pi*ln(2)$

Now by putting $k=12$ we get that $I(12)=\boxed{5.88}$