For the first part of the question I can calculate the final answer as $$\frac{2\tan^{-1}\sqrt{\frac{y-1}{y+1}}}{\sqrt{y^2-1}}$$ However I am still unable to see how to get the second part. I tried differentiating the first one with respect to $y$ but not sure how to go from there.
$$ I(y) = \int_0^{\pi/2}\frac{1}{y+\cos(x)}dx $$ we then have (by the fundamental theorem of Calculus) $$ \frac{dI}{dy} = \int_0^{\pi/2}\frac{d}{dy}\frac{1}{y+\cos(x)}dx = -\int_0^{\pi/2}\frac{1}{(y+\cos(x))^2}dx $$ Which leaves us with $$ \int_0^{\pi/2}\frac{1}{(y+\cos(x))^2}dx = -\frac{dI}{dy}. $$ Thus take your initial solution and take the derivative.
There is a fly in the ointment for the first integral: $$ \int \frac{1}{\cos (x)+y} \, dx = \frac{2 \tanh ^{-1}\left(\frac{(1-y) }{\sqrt{1-y^2}}\tan \left(\frac{x}{2}\right)\right)}{\sqrt{1-y^2}} $$
As noted by @Chinny84, the final answer is $$ - \frac{d}{dy} \frac{2 \tanh ^{-1}\left(\frac{(1-y) }{\sqrt{1-y^2}}\tan \left(\frac{x}{2}\right)\right)}{\sqrt{1-y^2}} = % \frac{\left(y^2-1\right) \sin (x)-2 y \sqrt{1-y^2} (\cos (x)+y) \tanh ^{-1}\left(\frac{(y-1) }{\sqrt{1-y^2}}\tan \left(\frac{x}{2}\right)\right)}{\left(y^2-1\right)^2 (\cos (x)+y)} $$
