check that $\int_0^{\pi/2}\frac{1}{\sqrt{1-k^2sin^2\theta}}d\theta$ is equivalent to $-\log\sqrt{1-k^2}$ for $k\to1$. Please help check that $\int_0^{\pi/2}\frac{1}{\sqrt{1-k^2sin^2\theta}}d\theta$ is equivalent to $-\log\sqrt{1-k^2}$ for $k\to1$.

I really have no clue how to do this. Could anyone kindly help or provide some hint?
Thanks so much!
Since $k \rightarrow 1$ try working backwards simplifying the given expression so that $k$ is a value. To check it, use the method of solving improper integrals.

I could be wrong. I would change the expression so that the denominator would be
$\sqrt{1-1{\sin{x}}^2}$. Then solve the integral using whatever method is necessary. Then compare the results? 

Sorry this should be a comment rather than an answer.

P.S. I only took first year courses for Calc, so sorry in advance, if you think my explanation is not clear and lacking some other information..
Since $k \rightarrow 1$ try working backwards simplifying the given expression so that $k$ is a value. To check it, use the method of solving improper integrals.

I could be wrong. I would change the expression so that the denominator would be
$\sqrt{1-1{\sin{x}}^2}$. Then solve the integral using whatever method is necessary. Then compare the results? 

Sorry this should be a comment rather than an answer.

P.S. I only took first year courses for Calc, so sorry in advance, if you think my explanation is not clear and lacking some other information..