The different results of the definite integral by two methods

Question

\begin{array}{l} Find\ the\ value\ of\int _{\frac{\pi }{2}}^{\frac{3\pi }{2}}\sqrt{( 1+sin\theta )^{2} +cos( \theta )^{2}} d\theta \\ \\ Method\ 1\ :\ Substitution\\ \int \sqrt{( 1+sin\theta )^{2} +cos( \theta )^{2}} d\theta \\ =\int \sqrt{1+2sin\theta +sin^{2} \theta +cos^{2} \theta } d\theta \\ =\int \sqrt{2+2sin\theta } d\theta \\ =\sqrt{2}\int ( 1+sin\theta )^{\frac{1}{2}} d\theta \ \ \ \ \ \ \ \ \ Let\ t=1+sin\theta ,\ dt=cos\theta d\theta \\ =\sqrt{2}\int t^{\frac{1}{2}} \cdot \frac{dt}{cos\theta } \ \ \ cos\theta =\sqrt{1-( t-1)^{2}}\\ =\sqrt{2}\int t^{\frac{1}{2}} \cdot \left( 1-( t-1)^{2}\right)^{-\frac{1}{2}} dt\\ \\ =\sqrt{2}\int \left(\frac{1}{2-t}\right)^{\frac{1}{2}} dt\ \ \ \ \ \ \ \ \ \ \ \ \ \ Let\ v=2-t,\ dv=-dt\\ =-\sqrt{2}\int v^{-\frac{1}{2}} dv\\ =-2\sqrt{2}( 1-sin\theta )^{\frac{1}{2}} +C\\ \\ \int _{\frac{\pi }{2}}^{\frac{3\pi }{2}}\sqrt{( 1+sin\theta )^{2} +cos( \theta )^{2}} d\theta =-2\sqrt{2}( 1-sin\theta )^{\frac{1}{2}} \mid _{\frac{\pi }{2}}^{\frac{3\pi }{2}} =-4\\ \\ Method\ 2:\ \\ \\ \int \sqrt{( 1+sin\theta )^{2} +cos( \theta )^{2}} d\theta \\ =\sqrt{2}\int \sqrt{( 1+sin\theta )} d\theta \ \ \\ =\sqrt{2}\int \sqrt{\left( sin\left(\frac{\theta }{2}\right) +cos\left(\frac{\theta }{2}\right)\right)^{2}} d\theta \ \\ =\sqrt{2}\left(\int sin\left(\frac{\theta }{2}\right) d\theta +\int cos\left(\frac{\theta }{2}\right) d\theta \right)\\ =2\sqrt{2}\left( sin\left(\frac{\theta }{2}\right) -cos\left(\frac{\theta }{2}\right)\right) +C\\ \\ \int _{\frac{\pi }{2}}^{\frac{3\pi }{2}}\sqrt{( 1+sin\theta )^{2} +cos( \theta )^{2}} d\theta =2\sqrt{2}\left( sin\left(\frac{\theta }{2}\right) -cos\left(\frac{\theta }{2}\right)\right) \mid _{\frac{\pi }{2}}^{\frac{3\pi }{2}} =4 \end{array} I want to know what is wrong with Method 1.
I've spent a lot of time checking the steps. Please help!