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+sinθ has period T=2π (the function can be obtained as a vertical translation of the sine function in the plane of coordinates (θ,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=2π:
l=∫Ids where ds=√r2+(drdθ)2dθ
So we have to compute the derivative:
drdθ=ddθ(1+sinθ)=cosθ
and this implies
ds=√(1+sinθ)2+(cosθ)2dθ=√1+2sinθ+sin2θ+cos2θdθ
By the Pythagorean trigonometric identity cos2θ+sin2θ=1:
ds=√2+2sinθdθ=√2(1+sinθ)dθ
Integrating that square root is not easy... we'd better take advantage of trigonometric identities. We have √1+sinθ (multiplied by a numerical coefficient) and we would like to get rid of that square root. The half angle formula for cosine states that cos2(α2)=1+cosα2. We can use this with sine too, because if α=θ−π2 then cosα=cos(θ−π2)=sinθ. So we substitute in the formula and get
cos2(θ−π22)=1+cos(θ−π2)2
and then
2cos2(θ2−π4)=1+sinθ
By substitution in the last ds expression that we wrote
ds=√4cos2(θ2−π4)dθ=2∣∣∣cos(θ2−π4)∣∣∣dθ
Now we can integrate this on an interval of length 2π. 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(x2−π4)≥0⇔−π2+2kπ≤x2−π4≤π2+2kπ
for k any integer value, so we get that −π2π+4kπ≤x≤32π+4kπ, and in particular −π2π≤x≤32π.
In the end, we can integrate the arc length on I=[−π2,32π], since this is an interval of length 2π. We can remove the absolute value, since cos(x2−π4)≥0 on I.
∫32π−π22cos(θ2−π4)dθ=4∫32π−π212cos(θ2−π4)dθ=4sin(θ2−π4)∣32π−π2=4sin(π2)−4sin(−π2)=4+4=8
