<s>[INST] Closed form of $I(t) = \int_0^{\pi/2}\frac{\cos xdx}{\sqrt{\sin^2 x+ t\cos^2 x}}$ What is the closed form of
$$I(t) = \int_0^{\pi/2}\frac{\cos xdx}{\sqrt{\sin^2 x+ t\cos^2 x}}$$
I tired the change of variables $$\int_a^bf(x) dx= \int_a^bf(a+b-x) dx$$
but is was not fruitful. 
 [/INST] For $t > 1$
$$I(t) = \int_0^{\pi/2}\frac{\cos xdx}{\sqrt{\sin^2 x+ t\cos^2 x}} = \int_0^{1}\frac{d(\sin x)}{\sqrt{\sin^2 x(1- t)+ t }} = \dfrac{\arcsin\left(\frac{\sqrt{t-1}}{\sqrt{t}}\right)}{\sqrt{t-1}}$$.

For $0 < t < 1$, let $ u = \sin x$
$$I(t) = \int_0^1 \dfrac{du}{\sqrt{u^2(1 - t) + t}} = \dfrac{1}{\sqrt t}\int_0^1 \dfrac{du}{\sqrt {(u((1-t)/t)^{1/2})^2+1} } = \dfrac{1}{\sqrt {1-t}}\int^{\sqrt{\frac{1-t}{t}}}_0 \dfrac{dz}{\sqrt{z^2 +1}}$$
where $z = u \sqrt{\dfrac{1- t}{t}}$. 
So $$I(t) = \dfrac{1}{\sqrt{1-t}}\log\left|\sqrt{\dfrac 1t} + \sqrt{\frac{1-t}{t}}\right|$$.
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