Problem 49 - Calculus Early Transcendentals

Question

Evaluate the following integrals or state that they diverge. $$\int_{-2}^{2} \frac{d p}{\sqrt{4-p^{2}}}$$

Step-by-Step Solution

Answer

If it converges, what is its value? Answer: The integral converges, and its value is $$\pi$$.

Step 1: Identify the Trigonometric Substitution

In order to evaluate the integral, we need to find a suitable trigonometric substitution to convert the expression into a simpler form. Since it is of the form $$\int \frac{d p}{\sqrt{4-p^{2}}},$$ we can use the substitution $$p = 2\sin\theta.$$ The reason is that in this case, the expression $$4 - p^2$$ becomes $$4\cos^2\theta,$$ which simplifies the square root.

Step 2: Find the Differential Substitution

Now that we have chosen the substitution $$p = 2\sin\theta$$, we need to find the differential substitution. Differentiate with respect to $$\theta$$ to find $$dp$$: $$dp = 2\cos\theta d\theta.$$

Step 3: Change the Limits of Integration

Since we are changing the variable from $$p$$ to $$\theta$$, we need to find the new limits of integration. For the lower limit, we have $$p = -2 \Rightarrow -2 = 2\sin\theta \Rightarrow \sin\theta = -1 \Rightarrow \theta = -\frac{\pi}{2}.$$ For the upper limit, we have $$p = 2 \Rightarrow 2 = 2\sin\theta \Rightarrow \sin\theta = 1 \Rightarrow \theta = \frac{\pi}{2}.$$ Therefore, the new limits of integration are $$-\frac{\pi}{2}$$ and $$\frac{\pi}{2}$$.

Step 4: Substitute and Simplify the Integral

Now, we substitute $$p$$ and $$dp$$ into the integral and simplify the expression: $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2\cos\theta d\theta}{\sqrt{4-4\sin^2 \theta }} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2\cos\theta d\theta}{\sqrt{4(1-\sin^2 \theta )}} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2\cos\theta d\theta}{\sqrt{4\cos^2 \theta }}.$$ The square root simplifies as: $$\sqrt{4\cos^2\theta} = 2\cos\theta.$$ Our integral then becomes $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta.$$

Step 5: Evaluate the Integral

Evaluating the integral, we have $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta = \left[\theta \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \left(\frac{\pi}{2}\right) - \left(-\frac{\pi}{2}\right) = \pi.$$ Thus, the given integral converges and its value is $$\pi$$.