Evaluating Integral: $\int_0^{\pi/2} \frac{1}{y+\cos x}dx$

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How do you evaluate an integral such as:
\begin{equation}
\int_0^\frac{\pi}{2} \frac{1}{y+cosx} \, dx
\end{equation}
I was thinking whether to treat y as a constant and then integrate as such and be left with an arbitrary constant that is a function of y. This constant, f(y), should then disappear when evaluating the definite integral...?
 
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That depends on what you want to find out. Does y have some fixed relationship to x?
Usually, y will be a constant, and the definite integral will still depend on y in the same way the integral will give different results if you replace y by different real numbers.
 
I’ve attached my attempt at the question. Just wanted to know what you think? I’ve got a definite integral that is a function of y, I(y), and have used the substitution t=tan(x/2).
 

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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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