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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..

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Please, use the tex support in this site (rather than posting a image link which might rot), and choose a better title. You have been here 5 months and have asked 24 questions. You should know about these things. -1 till you fix it. –  Aryabhata Mar 18 at 5:08
    
You may be expected to compute the integral up to $2-\epsilon$, and then take a limit. You have been told what substitution to use. –  André Nicolas Mar 18 at 5:21
    
The integrand is the derivative of $\arcsin(\frac{x}{2})$ –  Semsem Mar 18 at 6:54
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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 $\tfrac{\pi}{2}$.

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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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