TITLE: How to solve $\int \sqrt{1+\sin x}\, dx$?
QUESTION [6 upvotes]: It's easy to get this:
$$\int \sqrt{1+\sin x}\, dx \\= \int \sqrt{ \sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}} + 2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}\,\, dx  \\ = \int \left | \sin{\frac{x}{2}} + \cos{\frac{x}{2}} \right |\, dx \\= \sqrt{2} \int \left | \sin{\left ( \frac{x}{2} + \frac{\pi}{4} \right )} \right |\, dx$$
So, in fact, my question is how to solve the integrate $\int \left | \sin x \right| dx $.
Or how to deal with the integrate when have absolute in it?

REPLY [6 votes]: As zkutch has given how to calculate with absolute value here is another approach
To avoid confusion its better to do this we have $$\int \frac{\cos x}{\cos x}\sqrt{1+\sin x}dx=\int \frac{\cos x dx}{\sqrt{1-\sin x}}=-2\sqrt{1-\sin x}+C$$
Here in above integral i had put $1-\sin x=t$ which gave answer easily