How do I evaluate the indefinite integral
#int(1+cos(x))^2dx# ?

Question

How do I evaluate the indefinite integral
 #int(1+cos(x))^2dx# ?

Christopher Allers · Accepted Answer

#int(1+cosx)^2dx=3/2x+2sinx+1/4sin2x+C# Let us look at some details.
By multiplying out the integrand,
#int(1+cosx)^2dx=int(1+2cosx+cos^2x) dx#
by #cos^2x={1+cos2x}/2#,
#=int(1+2cosx+{1+cos2x}/2)dx#
by cleaning us the integrand,
#=int(3/2+2cosx+1/2cos2x)dx#
by finding antiderivatives,
#=3/2x+2sinx+1/4sin2x+C#

Charles Anctil · Answer

undefined To evaluate the indefinite integral $$ \int (1+\cos(x))^2 \, dx $$, use the following steps:

1. Expand the square:
$$ (1+\cos(x))^2 = (1+\cos(x))(1+\cos(x)) $$
$$ = 1 + 2\cos(x) + \cos^2(x) $$

2. Rewrite $\cos^2(x)$ using the trigonometric identity:
$$ \cos^2(x) = \frac{1 + \cos(2x)}{2} $$

3. Substitute the expanded expression into the integral:
$$ \int (1+\cos(x))^2 \, dx = \int \left(1 + 2\cos(x) + \frac{1 + \cos(2x)}{2}\right) \, dx $$

4. Integrate each term separately:
$$ \int 1 \, dx = x $$
$$ \int 2\cos(x) \, dx = 2\sin(x) $$
$$ \int \frac{1 + \cos(2x)}{2} \, dx = \frac{1}{2}x + \frac{1}{4}\sin(2x) $$

5. Combine the results:
$$ \int (1+\cos(x))^2 \, dx = x + 2\sin(x) + \frac{1}{2}x + \frac{1}{4}\sin(2x) $$

6. Simplify the expression:
$$ \int (1+\cos(x))^2 \, dx = \frac{3}{2}x + 2\sin(x) + \frac{1}{4}\sin(2x) + C $$
Where $ C $ is the constant of integration.

How do I evaluate the indefinite integral #int(1+cos(x))^2dx# ?