Trigonometric Integral

The first step is to expand:

$\int (1 + \cos x)^3 \, dx = \int \cos^3x + 3\cos^2x + 3\cos x + 1 \, dx.$

Now, it looks like we can use product-to-sum formulas to obtain $\cos^2 x = \tfrac{1}{2} + \tfrac{1}{2} \cos (2x)$ and $\cos^3 x = \tfrac{3}{4} \cos x + \tfrac{1}{4} \cos (3x)$ . We can then plug these back into the integral:

$\int \cos^3x + 3\cos^2x + 3\cos x + 1 \, dx = \int \tfrac{1}{4} \cos (3x) + \tfrac{3}{2} \cos (2x) + \tfrac{15}{4} \cos x + \tfrac{5}{2} \, dx.$

And finally, we can integrate $\int \cos(nx) \, dx = \tfrac{1}{n} \sin(nx) + c$ for each integer $n > 0$ . So:

$\int \tfrac{1}{4} \cos (3x) + \tfrac{3}{2} \cos (2x) + \tfrac{15}{4} \cos x + \tfrac{5}{2} \, dx = \boxed{\tfrac{1}{12} \sin (3x) + \tfrac{3}{4} \sin (2x) + \tfrac{15}{4} \sin x + \tfrac{5}{2} x + c.}$

$\begin{aligned} I & = \int (1+\cos x)^3 dx \\ & = \int \left(1+2\cos^2 \frac x2 - 1 \right) dx \\ & = 8 \int \cos^6 \frac x2 \ dx \\ & = 8 \int \left(\frac{e^{\frac x2 i} + e^{-\frac x2 i}}2 \right)^6 dx \\ & = \frac 18 \int \left(e^{3xi} + 6e^{2xi} + 15e^{xi} + 20 + 15 e^{-xi} +6 e^{-2xi} + e^{-3xi} \right) dx \\ & = \frac 18 \left(\frac{e^{3xi}}{3i} + 6\frac{e^{2xi}}{2i} + 15\frac{e^{xi}}{i} + 20x - 15\frac{e^{-xi}}{i} - 6\frac{e^{-2xi}}{2i} - \frac{e^{-3xi}}{3i} \right) + C \\ & = \frac 14 \left(\frac 13 \sin 3x + 3 \sin 2x + 15\sin x + 10x \right) + C \\ & = \boxed{\dfrac 1{12} \sin 3x + \dfrac 34 \sin 2x + \dfrac{15}4 \sin x + \dfrac 52 x + C} \end{aligned}$