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..
 A: The limit of integration for which our integrand is undefined is $x = 2$. So, what we really need to compute is:
\begin{align}
\lim \limits_{t \to 2} \int_0^t\dfrac{dx}{\sqrt{4-x^2}}
\end{align}
First, let's worry about the integral. Consider the substitution $x=2\sin(\theta)$ (do a bit of trigonometry to see why this might be useful). So, $dx=2\cos(\theta)d\theta$. So, the integral we're dealing with is:
\begin{align}
\int \dfrac{2\cos(\theta)d\theta}{\sqrt{4-4\sin^2(\theta)}}\\
= \int \dfrac{2\cos(\theta)d\theta}{2\sqrt{1-\sin^2(\theta)}}\\
= \int \dfrac{\cos(\theta)d\theta}{\cos(\theta)}\\
= \int d\theta\\
= \theta + C
\end{align}
Now, we need to put things back in terms of $x$. Solving for $\theta$, we see that this is equal to $\sin^{-1}(\tfrac{x}{2}) + C$.
Now, back to what we were originally dealing with. We now have:
\begin{align}
\lim \limits_{t \to 2}  \left(\sin^{-1}(\tfrac{x}{2}) \Big|_0^t\right)\\
= \sin^{-1}(\tfrac{2}{2}) - \sin^{-1}(\tfrac{0}{2})\\
= \tfrac{\pi}{2} - 0 = \boxed{\tfrac{\pi}{2}}
\end{align}
We now see that this improper integral converges to the value $\tfrac{\pi}{2}$.
A: 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,
$$\int_0^2\frac{dx}{\sqrt{4 - x^2}} = \left[\sin^{-1}{\frac{x}{2}}\right]_0^2\\
= \frac{\pi}{2}$$
