Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.3 Exercises - Page 531: 67

$$\eqalign{ & \left( {\text{a}} \right)\frac{1}{2}{\sin ^2}x + C \cr & \left( {\text{b}} \right) - \frac{1}{2}{\cos ^2}x + C \cr & \left( {\text{c}} \right)\frac{1}{2}{\sin ^2}x + C \cr & \left( {\text{d}} \right) - \frac{1}{4}\cos 2x + C \cr} $$

$$\eqalign{ & \int {\sin x\cos x} dx \cr & \cr & \left( {\text{a}} \right){\text{Using the substitution }}u = \sin x \cr & u = \sin x,{\text{ }}du = \cos xdx \cr & \int {\sin x\cos x} dx = \int u du \cr & {\text{Integrating}} \cr & = \frac{1}{2}{u^2} + C \cr & {\text{Write in terms of }}x \cr & = \frac{1}{2}{\left( {\sin x} \right)^2} + C \cr & = \frac{1}{2}{\sin ^2}x + C \cr & \cr & \left( {\text{b}} \right){\text{Using the substitution }}u = \cos x \cr & u = \cos x,{\text{ }}du = - \sin xdx \cr & \int {\sin x\cos x} dx = - \int u du \cr & {\text{Integrating}} \cr & = - \frac{1}{2}{u^2} + C \cr & {\text{Write in terms of }}x \cr & = - \frac{1}{2}{\left( {\cos x} \right)^2} + C \cr & = - \frac{1}{2}{\cos ^2}x + C \cr & {\text{Using the pythagorean identity co}}{{\text{s}}^2}x + {\sin ^2}x = 1 \cr & = - \frac{1}{2}\left( {1 - {{\sin }^2}x} \right) + C \cr & = \frac{1}{2}{\sin ^2}x - \frac{1}{2} + C \cr & \cr & \left( {\text{c}} \right){\text{ Integrating by parts}} \cr & u = \sin x,{\text{ }}du = \cos xdx \cr & dv = \cos xdx,{\text{ }}v = \sin x \cr & {\text{Using the integration by parts formula}} \cr & \int {\sin x\cos x} dx = \left( {\sin x} \right)\left( {\sin x} \right) - \int {\sin x} \cos xdx \cr & \int {\sin x\cos x} dx = {\sin ^2}x - \int {\sin x} \cos xdx \cr & {\text{Solve for }}\int {\sin x} \cos xdx \cr & 2\int {\sin x\cos x} dx = {\sin ^2}x \cr & \int {\sin x\cos x} dx = \frac{1}{2}{\sin ^2}x + C,{\text{ }}C = 0 \cr & \cr & \left( {\text{d}} \right){\text{Using the identity }}\sin 2x = 2\sin x\cos x \cr & \int {\sin x} \cos xdx = \frac{1}{2}\int {\sin 2x} dx \cr & {\text{Integrating}} \cr & = \frac{1}{2}\left( { - \frac{1}{2}\cos 2x} \right) + C \cr & = - \frac{1}{4}\cos 2x + C \cr & {\text{Use the identity }}\cos 2x = 1 - 2{\sin ^2}x \cr & = - \frac{1}{4}\left( {1 - 2{{\sin }^2}x} \right) + C \cr & = - \frac{1}{4} + \frac{1}{2}{\sin ^2}x + C \cr & = \frac{1}{2}{\sin ^2}x - \frac{1}{4} + C \cr & \cr & {\text{The answers differ in the constants}} \cr} $$