## Calculus (3rd Edition)

$$\frac{1}{8}\left[\cos \theta \sin \theta-2 \cos ^{3} \theta \sin \theta+\theta\right]+C$$
Since \begin{aligned} \int \cos ^{2} \theta \sin ^{2} \theta d \theta &=\int \cos ^{2} \theta\left(1-\cos ^{2} \theta\right) d \theta \\ &=\int \cos ^{2} \theta-\cos ^{4} \theta d \theta \\ &=\int \cos ^{2} \theta d \theta-\int \cos ^{4} \theta d \theta \end{aligned} Use $$\int \cos ^{n} x d x=\frac{1}{n} \cos ^{n-1} x \sin x+\frac{n-1}{n} \int \cos ^{n-2} x d x$$ Then \begin{aligned} \int \cos ^{2} \theta d x &=\frac{1}{2} \cos ^{2-1} \theta \sin \theta+\frac{2-1}{2} \int \cos ^{2-2} \theta d \theta \\ &=\frac{1}{2} \cos \theta \sin \theta+\frac{1}{2} \int 1 d \theta \\ &=\frac{1}{2} \cos \theta \sin \theta+\frac{1}{2} \theta+C \end{aligned} and \begin{aligned} \int \cos ^{4} \theta d \theta &=\frac{1}{4} \cos ^{4-1} \theta \sin \theta+\frac{4-1}{4} \int \cos ^{4-2} \theta d \theta \\ &=\frac{1}{4} \cos ^{3} \theta \sin \theta+\frac{3}{4} \int \cos ^{2} \theta d \theta \\ &=\frac{1}{4} \cos ^{3} \theta \sin \theta+\frac{3}{4}\left[\frac{1}{2} \cos \theta \sin \theta+\frac{1}{2} \theta\right]+C \\ &=\frac{1}{4} \cos ^{3} \theta \sin \theta+\frac{3}{8} \cos \theta \sin \theta+\frac{3}{8} \theta+C \end{aligned} Hence \begin{aligned} \int \cos ^{2} \theta d \theta-\int \cos ^{4} \theta d \theta &=\frac{1}{2} \cos \theta \sin \theta+\frac{1}{2} \theta-\left[\frac{1}{4} \cos ^{3} \theta \sin \theta+\frac{3}{8} \cos \theta \sin \theta+\frac{3}{8} \theta\right]+C \\ &=\frac{1}{2} \cos \theta \sin \theta+\frac{1}{2} \theta-\frac{1}{4} \cos ^{3} \theta \sin \theta-\frac{3}{8} \cos \theta \sin \theta-\frac{3}{8} \theta+C \\ &=\frac{1}{8} \cos \theta \sin \theta-\frac{1}{4} \cos ^{3} \theta \sin \theta+\frac{1}{8} \theta+C \\ &=\frac{1}{8}\left[\cos \theta \sin \theta-2 \cos ^{3} \theta \sin \theta+\theta\right]+C \end{aligned}