If $I=\int e^{x} \cos ^{2} x d x$, find $I$ using double angle formula.

Step 1 ：Given the integral, \(I = \int e^x \cos^2 x dx\)

Step 2 ：Using the double angle formula, \(\cos^2 x = \frac{1 + \cos(2x)}{2}\)

Step 3 ：Substitute the double angle formula into the integral, \(I = \int e^x \frac{1 + \cos(2x)}{2} dx\)

Step 4 ：Separate the integral, \(I = \frac{1}{2} \int e^x dx + \frac{1}{2} \int e^x \cos(2x) dx\)

Step 5 ：Integrate the first part, \(\frac{1}{2} \int e^x dx = \frac{1}{2} e^x + C_1\)

Step 6 ：Use integration by parts for the second part, let \(u = e^x\) and \(dv = \cos(2x) dx\)

Step 7 ：Then, \(du = e^x dx\) and \(v = \frac{1}{2} \sin(2x)\)

Step 8 ：Apply integration by parts, \(\int u dv = uv - \int v du\)

Step 9 ：\(\frac{1}{2} \int e^x \cos(2x) dx = \frac{1}{2} e^x \sin(2x) - \frac{1}{2} \int e^x \sin(2x) dx\)

Step 10 ：Use integration by parts again for the remaining integral, let \(u = e^x\) and \(dv = \sin(2x) dx\)

Step 11 ：Then, \(du = e^x dx\) and \(v = -\frac{1}{4} \cos(2x)\)

Step 12 ：Apply integration by parts, \(\int u dv = uv - \int v du\)

Step 13 ：\(-\frac{1}{2} \int e^x \sin(2x) dx = -\frac{1}{4} e^x \cos(2x) + \frac{1}{4} \int e^x \cos(2x) dx\)

Step 14 ：Solve for the remaining integral, \(\frac{1}{2} \int e^x \cos(2x) dx = \frac{1}{3} e^x \sin(2x) - \frac{1}{6} e^x \cos(2x) + C_2\)

Step 15 ：Combine the results, \(I = \frac{1}{2} e^x + \frac{1}{3} e^x \sin(2x) - \frac{1}{6} e^x \cos(2x) + C\)

Step 16 ：\(\boxed{I = \frac{e^x}{2} + \frac{e^x \sin(2x)}{3} - \frac{e^x \cos(2x)}{6} + C}\)