How do you evaluate the integral #int 1/(sqrt(4-x^2)dx# from 0 to 2?

Question

Samantha Adkison · Accepted Answer

#pi/2#.

Knowing that, #int1/sqrt(a^2-x^2)dx=arc sin(x/a)+C#, we have, #int_0^2 1/sqrt(4-x^2)dx=[arc sin(x/2)]_0^2=arc sin1-arc sin0=pi/2#.

Emily Ammerman · Answer

undefined To evaluate the integral $ \int_0^2 \frac{1}{\sqrt{4-x^2}} \, dx$, we can use the substitution method. Let $ x = 2\sin(\theta) $, then $ dx = 2\cos(\theta) \, d\theta $ and the limits of integration become $ \theta = \sin^{-1}(0) = 0 $ and $ \theta = \sin^{-1}(1) = \frac{\pi}{2} $.

$$
\int_0^2 \frac{1}{\sqrt{4-x^2}} \, dx = \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{4-(2\sin(\theta))^2}} \cdot 2\cos(\theta) \, d\theta
$$

Simplify the expression under the square root and integrate:

$$
= \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{4-4\sin^2(\theta)}} \cdot 2\cos(\theta) \, d\theta
$$

$$
= \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{4\cos^2(\theta)}} \cdot 2\cos(\theta) \, d\theta
$$

$$
= \int_0^{\frac{\pi}{2}} \frac{1}{2\cos(\theta)} \cdot 2\cos(\theta) \, d\theta
$$

$$
= \int_0^{\frac{\pi}{2}} d\theta
$$

Now, integrating $d\theta$ over the interval $0$ to $\frac{\pi}{2}$ gives:

$$
= \left[\theta\right]_0^{\frac{\pi}{2}}
$$

$$
= \frac{\pi}{2} - 0
$$

$$
= \frac{\pi}{2}
$$

So, $ \int_0^2 \frac{1}{\sqrt{4-x^2}} \, dx = \frac{\pi}{2} $.