Chapter 8: Techniques of Integration - Section 8.4 - Trigonometric Substitutions - Exercises 8.4 - Page 467: 45

$$4{\sin ^{ - 1}}\left( {\frac{{\sqrt x }}{2}} \right) + \sqrt x \sqrt {4 - x} + C $$

$$\eqalign{ & \int {\sqrt {\frac{{4 - x}}{x}} } dx \cr & {\text{We use the hint }}{u^2} = x \cr & {u^2} = x,\,\,\,\,2udu = dx \cr & {\text{Then}}{\text{,}} \cr & \int {\sqrt {\frac{{4 - x}}{x}} } dx = \int {\sqrt {\frac{{4 - {u^2}}}{{{u^2}}}} } \left( {2udu} \right) \cr & {\text{Simplifying, we get:}} \cr & = \int {\frac{{\sqrt {4 - {u^2}} }}{u}} \left( {2udu} \right) \cr & = \int {2\sqrt {4 - {u^2}} } du \cr & {\text{Integrate with trigonometric substitution}} \cr & {\text{We set}}{\text{, }}u = 2\sin \theta,{\text{ }}du = 2\cos \theta d\theta \cr & \int {2\sqrt {4 - {u^2}} } du = 2\int {\sqrt {4 - {{\left( {2\sin \theta } \right)}^2}} } \left( {2\cos \theta } \right)d\theta \cr & = 4\int {\sqrt {4 - 4{{\sin }^2}\theta } } \cos \theta d\theta \cr & = 8\int {\sqrt {1 - {{\sin }^2}\theta } \cos \theta } d\theta \cr & {\text{Use the fundamental identity }}{\cos ^2}\theta + {\sin ^2}\theta = 1 \cr & = 8\int {\sqrt {{{\cos }^2}\theta } \cos \theta } d\theta \cr & = 8\int {{{\cos }^2}\theta } d\theta \cr & {\text{Use the identity co}}{{\text{s}}^2}\theta = \frac{1}{2}\left( {1 + \cos 2\theta } \right) \cr & = 8\int {\frac{1}{2}\left( {1 + \cos 2\theta } \right)} d\theta \cr & = 4\int {\left( {1 + \cos 2\theta } \right)} d\theta \cr & {\text{Integrating}} \cr & = 4\left( {\theta + \frac{1}{2}\sin 2\theta } \right) + C \cr & = 4\theta + 2\sin 2\theta + C \cr & = 4\theta + 2\left( {2\sin \theta \cos \theta } \right) + C \cr & = 4\theta + 4\sin \theta \cos \theta + C \cr & {\text{We have }}u = 2\sin \theta,\,\,\,{\text{then }}\theta = {\sin ^{ - 1}}\left( {\frac{u}{2}} \right){\text{ and cos}}\theta = \frac{{\sqrt {4 - {u^2}} }}{2} \cr & = 4{\sin ^{ - 1}}\left( {\frac{u}{2}} \right) + 4\left( {\frac{u}{2}} \right)\left( {\frac{{\sqrt {4 - {u^2}} }}{2}} \right) + C \cr & = 4{\sin ^{ - 1}}\left( {\frac{u}{2}} \right) + u\sqrt {4 - {u^2}} + C \cr & {\text{Write in terms of }}x{\text{; replace }}u = \sqrt x \cr & = 4{\sin ^{ - 1}}\left( {\frac{{\sqrt x }}{2}} \right) + \sqrt x \sqrt {4 - x} + C \cr} $$