## Calculus: Early Transcendentals (2nd Edition)

$$h\left( x \right) = \frac{1}{2}x - \frac{1}{4}\sin 2x + \frac{1}{2} + \frac{1}{4}\sin 2$$
\eqalign{ & h'\left( x \right) = {\sin ^2}x,{\text{ }}h\left( 1 \right) = 1 \cr & {\text{use the hint }}{\sin ^2}x = \frac{{1 - \cos 2x}}{2} \cr & h'\left( x \right) = {\sin ^2}x = \frac{{1 - \cos 2x}}{2} \cr & h\left( x \right) = \int {h'\left( x \right)} dx \cr & h\left( x \right) = \int {\left( {\frac{{1 - \cos 2x}}{2}} \right)} dx \cr & {\text{split the numerator}} \cr & h\left( x \right) = \int {\left( {\frac{1}{2} - \frac{{\cos 2x}}{2}} \right)} dx \cr & h\left( x \right) = \int {\frac{1}{2}} dx - \int {\frac{{\cos 2x}}{2}} dx \cr & h\left( x \right) = \int {\frac{1}{2}} dx - \frac{1}{4}\int {\cos 2x\left( 2 \right)} dx \cr & {\text{find the antiderivative }} \cr & h\left( x \right) = \frac{1}{2}x - \frac{1}{4}\sin 2x + C \cr & {\text{with }}h\left( 1 \right) = 1 \cr & 1 = \frac{1}{2}\left( 1 \right) - \frac{1}{4}\sin 2\left( 1 \right) + C \cr & 1 = \frac{1}{2} - \frac{1}{4}\sin 2 + C \cr & C = \frac{1}{2} + \frac{1}{4}\sin 2 \cr & then{\text{ }} \cr & h\left( x \right) = \frac{1}{2}x - \frac{1}{4}\sin 2x + \frac{1}{2} + \frac{1}{4}\sin 2 \cr}