<s>[INST] How can I solve this question? Compute the value of the following improper integral. If it is divergent, type "Diverges" or "D".
$$\int_0^2 \frac{dx}{\sqrt{4-x^2}}$$
Do I make $u= 4-x^2$ then $du= -2x \, dx$
Not exactly sure..
 [/INST] Consider the substitution $x = 2\sin\theta$. From this we have $dx = 2\cos\theta\cdot d\theta$. Substitute these in and see what happens:
$$\int\frac{2\cos\theta \cdot d\theta}{\sqrt{4 - 4\sin^2\theta}}\\\\
= \int\frac{2\cos\theta \cdot d\theta}{2\cos\theta}\\
= \int d\theta\\
= \theta + C\\
= \sin^{-1}{\frac{x}{2}} + C$$
Hence,
$$\int_0^2\frac{dx}{\sqrt{4 - x^2}} = \left[\sin^{-1}{\frac{x}{2}}\right]_0^2\\
= \frac{\pi}{2}$$
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