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