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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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  • $\begingroup$ Since $\sqrt[]{}$ is a multi-branch function, we simply agree on what branch of it we are using. $\endgroup$ – user8960 Feb 17 '17 at 22:19
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    $\begingroup$ @user8960 No, that is false. If it were just cosine, then the original integral could be negative. To Leonardo, you are correct. Good catch! $\endgroup$ – Simply Beautiful Art Feb 17 '17 at 22:20
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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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  • $\begingroup$ yes, I see the intervals of cosine and sine play a roles. Simply Beautiful Art gave a nice example. $\endgroup$ – Leonardo Feb 17 '17 at 23:43
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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: $$\int \sqrt{1-\sin(x)}~dx=\int \sqrt{2\sin^2\left(\frac{\pi}{4}-x\right)}~dx=\sqrt{2}\cdot \int \sin\left(\frac{\pi}{4}-x\right)dx=\sqrt{2}\cos\left(\frac{\pi}{4}-x\right)+c$$

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  • $\begingroup$ Is my edit correct? $\endgroup$ – projectilemotion May 13 '17 at 12:16
  • $\begingroup$ 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: $$\int \sqrt{1-\sin{x}}~dx=\int \sqrt{1-\cos\left(\frac{\pi}{2}-x\right)}~dx=\int \sqrt{2\sin^2\left(\frac{\pi}{4}-\color{red}{\frac{x}{2}}\right)}~dx$$ $\endgroup$ – projectilemotion May 13 '17 at 13:04
  • $\begingroup$ yes , sorry for the mistake $\endgroup$ – Youssef Khiari May 13 '17 at 21:34
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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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    $\begingroup$ Since when $\sqrt{y^2}=y$? $\endgroup$ – egreg Feb 17 '17 at 23:11
  • $\begingroup$ @egreg : there is no limit for integration ,but you are right $$\sqrt{y^2}=|y|$$ $\endgroup$ – Khosrotash Feb 17 '17 at 23:14
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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:

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

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