91Ó°ÊÓ

We will start by rewriting the integral as a product of two cosine functions: $$\int \cos^3{x} dx = \int (\cos^2{x}\cos{x}) dx$$

We will now use the reduction formula for cosine which states that \(\cos^2{x} = \frac{1+\cos{2x}}{2}\). So, we can rewrite the integral as: $$\int (\cos^2{x}\cos{x}) dx = \int\left(\frac{1+\cos{2x}}{2}\cdot\cos{x}\right)dx$$ Now, we will distribute the multiplication inside the integral: $$\int\left(\frac{1+\cos{2x}}{2}\cdot\cos{x}\right)dx = \frac{1}{2}\int\cos{x}dx + \frac{1}{2}\int\cos{x}\cos{2x}dx$$

We will now solve the remaining integrals. The integral on the left is straightforward: $$\frac{1}{2}\int\cos{x}dx = \frac{1}{2}\sin{x} + C_1$$ For the integral on the right, we will use the substitution method. Let \(u = 2x\), so \(\frac{1}{2}du = dx\). Then the integral becomes: $$\frac{1}{2}\int\cos{x}\cos{2x}dx = \frac{1}{4}\int\cos{\frac{u}{2}}\cos{u}du$$ Now, we use the product-to-sum formula for cosine to further simplify the integral: \(\cos{\frac{u}{2}}\cos{u} = \frac{1}{2}\left(\cos{\frac{3u}{2}} + \cos{\frac{u}{2}}\right)\). Substituting this into the integral, we obtain: $$\frac{1}{4}\int\cos{\frac{u}{2}}\cos{u}du = \frac{1}{8}\int\left(\cos{\frac{3u}{2}} + \cos{\frac{u}{2}}\right)du$$ Now, integrate each term separately: $$\frac{1}{8}\int\left(\cos{\frac{3u}{2}}\right)du + \frac{1}{8}\int\left(\cos{\frac{u}{2}}\right)du$$ $$= \frac{1}{12}\sin{\frac{3u}{2}} + \frac{1}{4}\sin{\frac{u}{2}} + C_2$$ Substitute back for \(x\): $$\frac{1}{12}\sin{3x} + \frac{1}{4}\sin{x} + C_2$$ Now, combine the two parts of the integral: $$\int \cos^3{x} dx = \frac{1}{2}\sin{x} + \frac{1}{12}\sin{3x} + \frac{1}{4}\sin{x} + C$$