<s>[INST] Best way to integrate this? $$ \int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{\sin^2(x)}{\sqrt{1+\cos(x)}}dx $$
What would be the best way to go about integrating and evaluating this integral?
 [/INST] Substitute $\cos x =t \implies -\sin x dx =dt $: $$\int_{0}^{0.5} \frac{\sqrt{1-t^2}}{\sqrt{1+t}} dt =\int_0^{0.5} \sqrt{1-t}dt = \left[ -\frac 23 (1-t)^{3/2} \right]_0^{0.5} = \frac 23-\frac 23 \cdot \frac{1}{2\sqrt 2} = \frac 23 -\frac{1}{3\sqrt 2}$$
Second way: $$\frac{\sin^2x}{\sqrt{1+\cos x}}= \frac{4\sin^2(x/2)\cos^2 (x/2) }{\sqrt 2 \cos(x/2)} = 2\sqrt 2 \sin^2(x/2) \cos(x/2) $$ Then you have something of the form $u^2 du$.
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