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I tried to solve this.But I end up with two types of antiderivative function.Then how to find the definite integral of this function? My results are -2(Sin x/2 -Cos x/2) and 2√(1+sinx).Which one should i use to find definite integral?

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$\sqrt{1-\sin x}=\sqrt{\dfrac{\cos^2 x}{1+\sin x}}=\dfrac{|\cos x|}{\sqrt{1+\sin x}}$. Let $1+\sin x=t.$

The antiderivative is therefore, $\pm \int\dfrac{dt}{\sqrt{t}}=\pm2\sqrt{1+\sin x}+c$,depending on the sign of $\cos x$.

As far as the definite integral is concerned, split the integral into appropriate ranges: where $\cos x$ is $+$ or $-$, so that you can you use one of $\pm2\sqrt{1+\sin x}$.

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    $\begingroup$ Instead of "sinx" type "\sin x" (don't forget the space) for a much nicer look. $\endgroup$
    – B. Goddard
    Dec 28, 2020 at 16:12

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