Chapter 7 - Section 7.6 - Integration Using Tables and Computer Algebra Systems - 7.6 Exercises - Page 528: 43

$\frac{3}{8}x + \frac{3}{8}\sin x\cos x + \frac{1}{4}\sin x{\cos ^3}x + C$

$$\eqalign{ & \int {{{\cos }^4}x} dx \cr & {\text{Integrate by using a CAS }}\left( {{\text{See image below}}} \right){\text{ we obtain:}} \cr & \int {{{\cos }^4}x} dx = \frac{1}{{32}}\left( {12x + 8\sin 2x + \sin 4x} \right) + C \cr & = \frac{3}{8}x + \frac{1}{4}\sin 2x + \frac{1}{{32}}\sin 4x + C \cr & {\text{Using the double angle formula sin2}}\theta = 2\sin \theta \cos \theta \cr & = \frac{3}{8}x + \frac{1}{4}\left( {2\sin x\cos x} \right) + \frac{1}{{32}}\left( {2\sin 2x\cos 2x} \right) + C \cr & = \frac{3}{8}x + \frac{1}{2}\sin x\cos x + \frac{1}{{16}}\left( {2\sin x\cos x} \right)\cos 2x + C \cr & = \frac{3}{8}x + \frac{1}{2}\sin x\cos x + \frac{1}{8}\sin x\cos x\cos 2x + C \cr & {\text{Using the double angle formula }}\cos 2\theta = 2{\cos ^2}\theta - 1 \cr & = \frac{3}{8}x + \frac{1}{2}\sin x\cos x + \frac{1}{8}\sin x\cos x\left( {2{{\cos }^2}x - 1} \right) + C \cr & {\text{Simplifying}} \cr & = \frac{3}{8}x + \frac{1}{2}\sin x\cos x + \frac{1}{8}\left( {2\sin x{{\cos }^3}x - \sin x\cos x} \right) + C \cr & = \frac{3}{8}x + \frac{1}{2}\sin x\cos x + \frac{1}{4}\sin x{\cos ^3}x - \frac{1}{8}\sin x\cos x + C \cr & = \frac{3}{8}x + \frac{3}{8}\sin x\cos x + \frac{1}{4}\sin x{\cos ^3}x + C \cr & \cr & {\text{Integrating by tables using the entry 74}} \cr & \int {{{\cos }^n}x} dx = \frac{1}{n}{\cos ^{n - 1}}x\sin x + \frac{{n - 1}}{n}\int {{{\cos }^{n - 2}}x} dx \cr & {\text{Let }}n = 4 \cr & \int {{{\cos }^4}x} dx = \frac{1}{4}{\cos ^{4 - 1}}x\sin x + \frac{{4 - 1}}{4}\int {{{\cos }^{4 - 2}}x} dx \cr & \int {{{\cos }^4}x} dx = \frac{1}{4}{\cos ^3}x\sin x + \frac{3}{4}\int {{{\cos }^2}x} dx \cr & {\text{Use the entry 64}} \cr & \int {{{\cos }^4}x} dx = \frac{1}{4}{\cos ^3}x\sin x + \frac{3}{4}\left( {\frac{1}{2}x + \frac{1}{4}\sin 2x} \right) + C \cr & \int {{{\cos }^4}x} dx = \frac{1}{4}{\cos ^3}x\sin x + \frac{3}{8}x + \frac{3}{{16}}\sin 2x + C \cr & {\text{Use the double angle formula }}\sin 2x = 2\sin x\cos x \cr & \int {{{\cos }^4}x} dx = \frac{1}{4}{\cos ^3}x\sin x + \frac{3}{8}x + \frac{3}{8}\sin x\cos x + C \cr & {\text{Rearranging}} \cr & \int {{{\cos }^4}x} dx = \frac{3}{8}x + \frac{3}{8}\sin x\cos x + \frac{1}{4}\sin x{\cos ^3}x + C \cr & \cr & {\text{We obtain the same result}} \cr} $$