identities
$\displaystyle \cos(A + B) = {\color{red}\cos A \cos B} - \sin A \sin B$
$\displaystyle \cos(A - B) = {\color{red} \cos A \cos B} + \sin A \sin B$
Add the equations together.
If A = B:
$\displaystyle \cos (2A) = \cos(A + A) = \cos A \cos A - \sin A \sin A = \cos^{2} A - \sin^{2} A$
Convert into terms of only cosA and you should get your answer
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Use your result from the first question to split your products in to sums. It'll be easier to integrate.
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$\displaystyle x = \cos t \: \: \Rightarrow \: \: dx = -\sin t dt$
Make the subs:
$\displaystyle \int \frac{\cos^{2} t}{\sqrt{1 - \cos^{2} t}} \: \cdot -\sin t dt \: \: = \: \:- \int \frac{\cos^{2} t \sin t}{\sqrt{\sin^{2} t}} dt$
etc.
You'll need your half-angle identity from your first question to do it after you simplify.
For the third part $\displaystyle \int\frac{x^2}{\sqrt{1-x^2}}$
let $\displaystyle x=\cos(\theta)$...dont often see this sub...
then $\displaystyle dx=-\sin(\theta)$
and $\displaystyle x^2=\cos^2(\theta)$
we need to find the limits of integration so $\displaystyle \frac{1}{2}=\cos(\theta)\Rightarrow(\theta=\frac{\ pi}{3}$
and $\displaystyle 0=\cos(\theta)\Rightarrow{\theta=\frac{\pi}{2}}$
So putting that in we have
$\displaystyle \int_{\frac{\pi}{2}}^{\frac{\pi}{3}}\frac{\cos^2(\ theta)\cdot{-\sin(\theta)}d\theta}{\sqrt{1-\cos^2(\theta)}}$$\displaystyle =\int_{\frac{\pi}{2}}^{\frac{\pi}{3}}\frac{\cos^2( \theta)-\sin(\theta)d\theta}{\sin(\theta)}$
Simplifying we get $\displaystyle -\int_{\frac{\pi}{2}}^{\frac{\pi}{3}}\cos^2(\theta) d\theta$
Using the identity $\displaystyle \cos^2(\theta)=\frac{1+\cos(2\theta)}{2}$
we see this is equal to $\displaystyle -\bigg[\frac{1}{2}\sin(2\theta)+\frac{\theta}{2}\bigg|_{\ frac{\pi}{2}}^{\frac{\pi}{3}}=.045293$