Vtyjkyuk

Find the length of $r=1+sintheta$.

I got to $sqrt2 intlimits_0^2pisqrt1+sintheta ,mathrmdtheta$.

And the first way I used to solve the integral was substitution of $1+sintheta=u$.

Thus $ costheta ,mathrmdtheta = mathrmdu$. and I used $costheta=+sqrt1-sin^2theta$
to find the interval which is $[-pi/2,pi/2]$. Considering the original interval of $[0,2pi]$, I multiplied the result by $2$ for the symmetry to make the length of the interval $2pi$, without knowing if it's correct.

The second way is $frac1+cosphi2=cos^2(phi/2)$, substituting $phi=theta-pi/2$,

thus $sqrt1+sintheta=sqrt2cos^2(theta/2-pi/4)=+sqrt2cos(theta/2-pi/4)$.

From $-pi/2 <theta/2-pi/4<pi/2$, I got the interval $[-pi/2,3pi/2]$.

I end up with an interval length of $2pi$, so I just found the final result from this interval.

So, I used two method to solve the integral and needed to reconsider the interval for integral.

Though I got a result, I'm not fully understanding the way to find the interval.

Could you tell why these methods are correct or incorrect?

Edit: I should have mentioned I found the answer of $8$. It's not the way to get the answer that I'm asking. I'm confused about interval.

Length of the curve is

$$I=int_0^2 pisqrtr^2+left(fracdrdthetaright)^2dtheta = int_0^2 pisqrtr^2+cos^2theta hspace3ptd theta$$
$$I=int_0^2 pisqrt1+2sin theta +sin^2theta+cos^2theta hspace3ptd theta$$
$$I=sqrt2int_0^2pisqrt1+sintheta;dtheta$$
$$I=sqrt2displaystyleint_0^2pisqrt2cosleft(dfrac2theta-pi4right),mathrmdtheta$$

$$I=2displaystyleint_0^2picosleft(dfrac2theta-pi4right),mathrmdtheta$$

$u=dfrac2theta-pi4$ thus $mathrmdtheta=2,mathrmdu$

$$I=4classsteps-nodedisplaystyleintcosleft(uright),mathrmdu=-4sin u+c=-4sin(dfrac2theta-pi4)+c$$

As $theta$ goes from 0 to $2 pi$

$$I=-4sin(dfrac3pi4)+4sin(dfrac-pi4)=-4sqrt2$$