what will be the value of this integral $$ \large{ \int^{\Large{\frac{\pi}{2}}}_{0} \left[ e^{\ln\left(\cos x \cdot \frac{d(\cos x)}{dx}\right)} \right]dx}$$
We know that $\large{a^{log_a(c)} = c}$.
But in this question, the expression in logarithm function evaluates to be negative, for which log is not defined.
$$ \large{ \int^{\Large{\frac{\pi}{2}}}_{0} \left[ e^{\ln\left(-\frac{sin(2x)}{2}\right)} \right]dx}$$
So this means we cannot apply the log property? my working:
$$\large{= \int^{\Large{\frac{\pi}{2}}}_{0}\left[\cos x \cdot (- \sin x) \right]} dx$$
$$=\left[\dfrac{\cos^2 x}{2}\right]^{\frac{\pi}{2}}_{0}$$
$$=\boxed{-\dfrac{1}{2}}$$
Is this integral, with undefined log, correct?
 A: The question is: Which logarithm function are you using?
If you are using the real-valued $\ln \colon (0,+\infty) \to \mathbb{R}$, your integrand is simply not defined on the interval $\bigl[0,\frac{\pi}{2}\bigr]$, and hence the integral neither.

I put it into Alpha and it gave me -1/2 as answer.

Unless you tell it not to, Wolfram Alpha happily uses a branch of the complex logarithm, and you only have definition problems for the integrand at $0$ and $\frac{\pi}{2}$, where $\cos x = 0$ or $\frac{d}{dx}\cos x = 0$. At all other points, the integrand is well-defined - the ambiguity of $\ln$ is cancelled by the periodicity of $\exp$, so we don't even need to choose a consistent branch of the logarithm, we might choose a branch for any point independently of the other choices. Then you have, by definition of the logarithm
$$\exp \biggl(\ln \Bigl(\cos x \frac{d(\cos x)}{dx}\Bigr)\biggr) = \cos x \frac{d(\cos x)}{dx} = \frac{1}{2}\cdot \frac{d(\cos^2 x)}{dx},$$
and the integral evaluates to $-\frac{1}{2}$ regardless of how we define the integrand at the endpoints of the interval.
A: Even integrals can be undefined. In your case there will be no curve drawn for$$f(x)=e^{- \frac12sin2x}$$ so the area is also undefined.
