Question

problem 1 (int 4(1 + cos(x))^2 dx)

Explanation:

Step1: Expand the square

First, we expand \((1 + \cos(x))^2\) using the formula \((a + b)^2 = a^2 + 2ab + b^2\). Here, \(a = 1\) and \(b=\cos(x)\), so \((1 + \cos(x))^2=1 + 2\cos(x)+\cos^2(x)\). Then the integrand becomes \(4(1 + 2\cos(x)+\cos^2(x)) = 4 + 8\cos(x)+4\cos^2(x)\).

Step2: Use the double - angle formula for \(\cos^2(x)\)

Recall the double - angle formula \(\cos^2(x)=\frac{1 + \cos(2x)}{2}\). Substitute this into the integrand:
\(4 + 8\cos(x)+4\times\frac{1+\cos(2x)}{2}=4 + 8\cos(x)+2(1 + \cos(2x))\)
Simplify the expression: \(4 + 8\cos(x)+2 + 2\cos(2x)=6 + 8\cos(x)+2\cos(2x)\)

Step3: Integrate term - by - term

We know that \(\int kdx=kx + C\) (where \(k\) is a constant), \(\int\cos(x)dx=\sin(x)+C\), and \(\int\cos(ax)dx=\frac{1}{a}\sin(ax)+C\) (\(a
eq0\)).
Integrate \(6 + 8\cos(x)+2\cos(2x)\) with respect to \(x\):
\(\int(6 + 8\cos(x)+2\cos(2x))dx=\int6dx+\int8\cos(x)dx+\int2\cos(2x)dx\)
\(\int6dx = 6x\)
\(\int8\cos(x)dx=8\sin(x)\)
For \(\int2\cos(2x)dx\), let \(u = 2x\), then \(du=2dx\), and \(\int2\cos(2x)dx=\int\cos(u)du=\sin(u)+C=\sin(2x)+C\)
Combining these results, we get \(6x + 8\sin(x)+\sin(2x)+C\) (where \(C\) is the constant of integration)

Answer:

\(6x + 8\sin(x)+\sin(2x)+C\) (or we can also write \(\sin(2x)\) as \(2\sin(x)\cos(x)\), so the answer can also be written as \(6x + 8\sin(x)+2\sin(x)\cos(x)+C\))