How do you evaluate #int (1/(sqrt(x)sqrt(4-x)))dx# for [1, 2]?

Question

Scarlett Allison · Accepted Answer

#pi/6# #int_1^2 1/(sqrtx sqrt(4-x))dx=int_1^2 dx/sqrt(4x-x^2)=# #=int_1^2 dx/sqrt(4-4+4x-x^2)=int_1^2 dx/sqrt(4-(x^2-4x+4))=# #=int_1^2 dx/sqrt(4-(x-2)^2)=1/2int_1^2 dx/sqrt(1-((x-2)/2)^2)=I# #(x-2)/2=t => dx=2dt# #x=1 => t=-1/2# #x=2 => t=0# #I=int_(-1/2)^0 dt/sqrt(1-t^2)=arcsint|_(-1/2)^0# #I=0-(-pi/6)=pi/6#

Luke Alvarez · Answer

undefined To evaluate the integral $\int_{1}^{2} \frac{1}{\sqrt{x}\sqrt{4-x}} \, dx$, we can use the substitution method. Let $u = \sqrt{x}$. Then, $x = u^2$ and $dx = 2u \, du$. Substituting these into the integral, we get:

$$
\int_{1}^{2} \frac{1}{\sqrt{x}\sqrt{4-x}} \, dx = \int_{\sqrt{1}}^{\sqrt{2}} \frac{1}{u\sqrt{4-u^2}} \cdot 2u \, du
$$

$$
= 2\int_{1}^{\sqrt{2}} \frac{1}{\sqrt{4-u^2}} \, du
$$

Now, we can use a trigonometric substitution. Let $u = 2\sin(\theta)$, then $du = 2\cos(\theta) \, d\theta$ and $\sqrt{4-u^2} = 2\cos(\theta)$. Substituting these, we get:

$$
2\int_{1}^{\sqrt{2}} \frac{1}{2\cos(\theta)} \cdot 2\cos(\theta) \, d\theta = 2\int_{\sin^{-1}(1/2)}^{\sin^{-1}(1/\sqrt{2})} \, d\theta
$$

$$
= 2\left[\theta\right]_{\sin^{-1}(1/2)}^{\sin^{-1}(1/\sqrt{2})} = 2\left[\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) - \sin^{-1}\left(\frac{1}{2}\right)\right]
$$

$$
= 2\left[\frac{\pi}{4} - \frac{\pi}{6}\right] = 2\left(\frac{\pi}{4} - \frac{\pi}{6}\right) = 2\left(\frac{\pi}{12}\right) = \frac{\pi}{6}
$$

So, $\int_{1}^{2} \frac{1}{\sqrt{x}\sqrt{4-x}} \, dx = \frac{\pi}{6}$.