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One of the method to find the integral $$\int\sqrt{1-\sin x}\ dx$$ is by multiplying by $\dfrac{1+\sin x}{1+\sin x}$ inside the root. Then, by using the identity $\sin^2x+\cos^2x=1$ , we get $$\int\dfrac{\sqrt{\cos^2x}}{\sqrt{1+\sin x}}\ dx$$

The next step is we remove the square with the root and using the substitution $u=\sin x$. My question is why ? Why don't we put an absolute value of $\cos x$? So, we have two answers. Is this situation always true in any similar situation in indefinite integrals?

Sorry, if my question is trivial. Thanks

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    Since $\sqrt[]{}$ is a multi-branch function, we simply agree on what branch of it we are using. – 2017-02-17
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    @user8960 No, that is false. If it were just cosine, then the original integral could be negative. To Leonardo, you are correct. Good catch! – 2017-02-17

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As Simply Beautiful Art says, you are right. The indefinite integral obtained by using $\sqrt{\cos(x)^2} = \cos x$ is only valid between $\pm \frac{\pi}{2}$ (and at $2\pi$ intervals).

The actual value of the integral (praise Wolfram with your whole heart) is $$\frac{2 \sqrt{1-\sin (x)} \left(\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)\right)}{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}$$

WolframAlpha's "show steps" functionality makes the same mistake that you pointed out, and then cops out by saying at the end "This is equivalent, for restricted values of $x$, to the actual answer". Of course, Mathematica's Integrate function gets it right because it uses voodoo.

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    yes, I see the intervals of cosine and sine play a roles. Simply Beautiful Art gave a nice example. – 2017-02-17
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Another way: \begin{align} \int\sqrt{1-\sin x}\ dx&=\int\sqrt{1-2\sin \dfrac x2 \cos \dfrac x2}\ dx\\ &=\int\sqrt{\left(\sin^2 \dfrac x2+\cos^2 \dfrac x2\right)-2\sin \dfrac x2 \cos \dfrac x2}\ dx\\ &=\int\sqrt{\left(\sin \dfrac x2-\cos \dfrac x2\right)^2}\ dx\\ &=\int \left(\sin \dfrac x2-\cos \dfrac x2\right)\ dx\\ &=2\left(-\cos\dfrac x2-\sin\dfrac x2\right)+\text{const.} \end{align}

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    Since when $\sqrt{y^2}=y$? – 2017-02-17
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    @egreg : there is no limit for integration ,but you are right $$\sqrt{y^2}=|y|$$ – 2017-02-17
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Here is another way: $$\int \sqrt{1-\sin(x)}~dx=\int \sqrt{1-\cos\left(\frac{\pi}{2}-x\right)}~dx$$ We know that: $$1-\cos(\theta)\equiv 2\sin^2\left(\frac{\theta}{2}\right)$$ Hence: $$\int \sqrt{1-\sin(x)}~dx=\int \sqrt{2\sin^2\left(\frac{\pi}{4}-x\right)}~dx=\sqrt{2}\cdot \int \sin\left(\frac{\pi}{4}-x\right)=\sqrt{2}\cos\left(\frac{\pi}{4}-x\right)+c$$

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    Is my edit correct? – 2017-05-13
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    On the other hand, I am quite sure that your answer is incorrect. Note that if you apply: $$1-\cos(\theta)\equiv 2\sin^2\left(\frac{\theta}{2}\right)$$ And then you let $\theta=\frac{\pi}{2}-x$, then you obtain: $$\int \sqrt{1-\sin{x}}~dx=\int \sqrt{1-\cos\left(\frac{\pi}{2}-x\right)}~dx=\int \sqrt{2\sin^2\left(\frac{\pi}{4}-\color{red}{\frac{x}{2}}\right)}~dx$$ – 2017-05-13
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    yes , sorry for the mistake – 2017-05-13
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I would like to suggest another solution, using the substitution $sin(x) = sin^2(u)$. Then $x = arcsin(sin^2(u))$ (note that $arcsin(t) = sin^-1(t)$, $arcsin(t)$ is the inverse function of $sin(t)$).

Hence, $dx = d \biggl( arcsin(sin^2(u))\biggr) \Rightarrow dx = \frac{d\biggl(sin^2(u) \biggr)}{\sqrt{1-\bigl(sin^2(u)\bigr)^2}}$

$$\int \sqrt{1-sin(x)}dx = \int \sqrt{1 - sin^2(u)}\frac{d\biggl(sin^2(u) \biggr)}{\sqrt{1-\bigl(sin^2(u)\bigr)^2}}$$

$$\sqrt{1-\bigl(sin^2(u)\bigr)^2} = \sqrt{(1 - sin^2(u))(1 + sin^2(u))}$$

$$\int \frac{d\biggl(sin^2(u) \biggr)}{\sqrt{1+sin^2(u)}} = \int \frac{d\biggl(1 +sin^2(u) \biggr)}{\sqrt{1+sin^2(u)}}$$
Using the formula: $\int \frac{du}{\sqrt u} = 2u^{1/2} + C$ we get:

$$2\biggl(1 + sin^2(u)\biggr)^{1/2} + C$$ and the final result is:

$$\int \sqrt{1-sin(x)}dx = 2\biggl(1 + sin^2(u)\biggr)^{1/2} + C = 2\biggl(1 + sin(x)\biggr)^{1/2} + C.$$