This is the equation of a *cardioid*. The polar equation is expressed in terms of a periodic function, so we have to know its period to get the exact length (and not just a portion or some excess of it).

#1+sin theta# has period #T=2pi# (the function can be obtained as a vertical translation of the sine function in the plane of coordinates #(theta,r)#).

The length of a periodic polar curve can be computed by integrating the arc length on a complete period of the function, i.e. on an interval #I# of length #T=2 pi#:

#l=int_{I}ds# where #ds=sqrt(r^2+((dr)/(d theta))^2) d theta#

So we have to compute the derivative:

#(dr)/(d theta)=d/(d theta) (1 + sin theta)=cos theta #

and this implies

#ds=sqrt((1+sin theta )^2+(cos theta )^2)d theta=sqrt(1+2sin theta +sin^2 theta +cos^2 theta ) d theta#

By the Pythagorean trigonometric identity #cos^2 theta +sin^2 theta =1#:

#ds=sqrt(2+2sin theta )d theta=sqrt(2(1+sin theta)) d theta#

Integrating that square root is not easy... we'd better take advantage of trigonometric identities. We have #sqrt(1+sin theta)# (multiplied by a numerical coefficient) and we would like to get rid of that square root. The *half angle formula* for cosine states that #cos^2(alpha/ 2)= (1+cos alpha)/2#. We can use this with sine too, because if #alpha=theta-pi/2# then #cos alpha = cos(theta - pi/2) = sin theta#. So we substitute in the formula and get

#cos^2((theta-pi/2)/2)=(1+cos(theta - pi/2))/2#

and then

#2cos^2(theta/2-pi/4)=1+sin theta#

By substitution in the last #ds# expression that we wrote

#ds=sqrt(4cos^2(theta/2 - pi/4)) d theta=2|cos(theta/2 - pi/4)| d theta#

Now we can integrate this on an interval of length #2pi#. To do that, we would like to get rid of the absolute value. The graph of the integrand function is the following

graph{2*abs(cos(x/2-pi/4)) [-10, 10, -5, 5]}

We know that

#cos(x/2-pi/4) ge 0 <=> -pi/2+2k pi le x/2-pi/4 le pi/2 + 2k pi#

for #k# any integer value, so we get that #- pi/2 pi+4k pi le x le 3/2 pi + 4k pi#, and in particular #- pi/2 pi le x le 3/2 pi#.

In the end, we can integrate the arc length on #I=[-pi/2,3/2 pi]#, since this is an interval of length #2pi#. We can remove the absolute value, since #cos(x/2-pi/4) ge 0 # on #I#.

#int_{-pi/2}^{3/2 pi} 2cos(theta/2 - pi/4) d theta=4 int_{-pi/2}^{3/2 pi} 1/2 cos(theta/2 - pi/4) d theta=4 sin(theta/2 - pi/4) |_{-pi/2}^{3/2 pi} = 4 sin(pi/2)-4sin(-pi/2)=4+4=8#