ANSWER
[tex]-2\sqrt[]{1+\cos(x)}+C[/tex]
EXPLANATION
To solve this integral we have to use the substitution method. Let u = 1 + cos(x), then du is,
[tex]du=-\sin (x)dx[/tex]
Thus, dx is,
[tex]dx=\frac{du}{-\sin (x)}[/tex]
Replace the function and the differential in the integral,
[tex]\int \frac{\sin(x)}{\sqrt[]{1+\cos(x)}}dx=\int \frac{\sin(x)}{\sqrt[]{u}}\cdot\frac{du}{-\sin (x)}[/tex]
The sin(x) cancels out,
[tex]\int \frac{\sin(x)}{\sqrt[]{u}}\cdot\frac{du}{-\sin(x)}=-\int \frac{1}{\sqrt[]{u}}du[/tex]
We have to find a function whose derivative is 1/√u. This function is √u since its derivative is,
[tex]\frac{d}{du}(\sqrt[]{u})=\frac{1}{2\sqrt[]{u}}[/tex]
Note that a coefficient 1/2 is missing, so to cancel it out, we have to multiply by 2. Don't forget the constant of integration,
[tex]-\int \frac{1}{\sqrt[]{u}}du=-2\sqrt[]{u}+C[/tex]
Finally, we have to replace u with the function we substituted before,
[tex]-2\sqrt[]{u}+C=-2\sqrt[]{1+\cos (x)}+C[/tex]
Hence, the result of the integral is,
[tex]-2\sqrt[]{1+\cos(x)}+C[/tex]
