What is the area bounded by the the inside of polar curve #1+cos theta# and outside the polar curve #r(1+cos theta)=1#?

If we plot the polar curve, and shadethe area sought:

We observe that the points of intersection are:

#theta = -pi/2# and #theta=pi/2#

We calculate area in polar coordinates using :

# A = 1/2 \ int_alpha^beta \ r^2 \ d theta #

Thus, the enclosed area is:

# A = 1/2 \ int_(-pi/2)^(pi/2) \ (1+cos theta)^2 - (1/(1+cos theta))^2 \ d theta #

# :. 2A = int_(-pi/2)^(pi/2) \ (1+cos theta)^2 \ d theta - int_(-pi/2)^(pi/2) (1/(1+cos theta))^2 \ d theta #

Consider the first integral;:

# I_1= int_(-pi/2)^(pi/2) \ (1+cos theta)^2 \ d theta #
# \ \ \ = int_(-pi/2)^(pi/2) \ 1+ 2cos theta + cos^2 theta \ d theta #
# \ \ \ = int_(-pi/2)^(pi/2) \ 1+ 2cos theta + (1+cos 2theta)/2 \ d theta #
# \ \ \ = int_(-pi/2)^(pi/2) \ 3/2+ 2cos theta + (cos 2theta)/2 \ d theta #
# \ \ \ = [(3theta)/2+ 2sin theta + (sin 2theta)/4]_(-pi/2)^(pi/2) \ #

# \ \ \ = ((3(pi/2))/2+ 2sin (pi/2)+ (sin (2pi))/4) - ((3(-pi/2))/2+ 2sin (-pi/2)+ (sin (-2pi))/4)#

# \ \ \ = ((3pi)/4+ 2+ 0) - (-(3pi)/4-2-0)#
# \ \ \ = (3pi)/2+ 4#

And, now the second integral:

# I_2 = int_(-pi/2)^(pi/2) (1/(1+cos theta))^2 \ d theta #

For which, we perform, a tangent half-angle substitution:

# u =tan(theta/2) => (du)/(d theta) = 1/2sec^2(theta/2) #

When:

# theta = { (pi/2), (-pi/2) :} => u = { (1), (-1) :} #

And we can manipulate the integral and perform the substitution, to get:

# I_2 = int_(-pi/2)^(pi/2) 1/( (1-tan^2(theta/2))/(1+tan^2(theta/2))+1)^2 \ d theta #
# \ \ \ = int_(-pi/2)^(pi/2) 1/( (1-tan^2(theta/2))/(sec^2(theta/2))+1)^2 \ d theta #
# \ \ \ = int_(-pi/2)^(pi/2) 1/ ((1-tan^2(theta/2)+1+tan^2(theta/2))/(1+tan^2(theta/2)))^2 \ d theta #
# \ \ \ = int_(-pi/2)^(pi/2) 1/ ((2)/(sec^2(theta/2)))^2 \ d theta #
# \ \ \ = int_(-pi/2)^(pi/2) 1/ ((2)/(sec^2(theta/2)))^2 \ d theta #
# \ \ \ = 1/4 \ int_(-pi/2)^(pi/2) sec^2 (theta/2) sec^2 (theta/2) \ d theta #
# \ \ \ = 1/2 \ int_(-pi/2)^(pi/2) (1+tan^2 (theta/2)) \ 1/2sec^2 (theta/2) \ d theta #
# \ \ \ = 1/2 \ int_(-1)^(1) 1+u^2 \ du #
# \ \ \ = 1/2 [u+u^3/3]_(-1)^(1) #
# \ \ \ = 1/2 {(1+1/3) - (-1-1/3)} #
# \ \ \ = 4/3 #

Combining both results, we get:

# 2A = I_1 + I_2 #
# \ \ \ \ \ = (3pi)/2+ 4 + 4/3 #
# \ \ \ \ \ = (3pi)/2+ 16/3 #

Hence:

# A = (3pi)/4+ 4/3 #