Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises - Page 393: 102

$$\eqalign{ & {\text{Proved:}} \cr & \int {{{\sin }^2}ax} dx = \frac{x}{2} - \frac{{\sin \left( {2ax} \right)}}{{4a}} + C \cr & {\text{ and }} \cr & \int {{{\cos }^2}ax} dx = \frac{x}{2} + \frac{{\sin \left( {2ax} \right)}}{{4a}} + C \cr} $$

$$\eqalign{ & \int {{{\sin }^2}ax} dx \cr & {\text{We need to use the identity si}}{{\text{n}}^2}\theta = \frac{{1 - \cos 2\theta }}{2}, \cr & \int {{{\sin }^2}ax} dx = \int {\frac{{1 - \cos \left( {2ax} \right)}}{2}} dx \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int {\left( {\frac{1}{2} - \frac{1}{2}\cos \left( {2ax} \right)} \right)} dx \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\int {dx} - \frac{1}{2}\int {\cos \left( {2ax} \right)} dx \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{2} - \frac{1}{2}\int {\cos \left( {2ax} \right)} dx \cr & {\text{Let }}u = 2ax,\,\,\,du = 2adx \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{2} - \frac{1}{2}\int {\cos u\left( {\frac{1}{{2a}}} \right)} du \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{2} - \frac{1}{{4a}}\int {\cos u} du \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{2} - \frac{{\sin u}}{{4a}} + C \cr & {\text{Replace back }}u = 2ax \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{2} - \frac{{\sin \left( {2ax} \right)}}{{4a}} + C \cr & \cr & \int {{{\cos }^2}ax} dx \cr & {\text{We need to use the identity }}{\cos ^2}\theta = \frac{{1 + \cos 2\theta }}{2}, \cr & \int {{{\cos }^2}ax} dx = \int {\frac{{1 + \cos \left( {2ax} \right)}}{2}} dx \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int {\left( {\frac{1}{2} + \frac{1}{2}\cos \left( {2ax} \right)} \right)} dx \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\int {dx} + \frac{1}{2}\int {\cos \left( {2ax} \right)} dx \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{2} + \frac{1}{2}\int {\cos \left( {2ax} \right)} dx \cr & {\text{Let }}u = 2ax,\,\,\,du = 2adx \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{2} + \frac{1}{2}\int {\cos u\left( {\frac{1}{{2a}}} \right)} du \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{2} + \frac{1}{{4a}}\int {\cos u} du \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{2} + \frac{{\sin u}}{{4a}} + C \cr & {\text{Replace back }}u = 2ax \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{2} + \frac{{\sin \left( {2ax} \right)}}{{4a}} + C \cr} $$