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

  • $\begingroup$ 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
  • 5
    $\begingroup$ @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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

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: $$\int \sqrt{1-\sin(x)}~dx=\int \sqrt{2\sin^2\left(\frac{\pi}{4}-x\right)}~dx=\sqrt{2}\cdot \int \sin\left(\frac{\pi}{4}-x\right)dx=\sqrt{2}\cos\left(\frac{\pi}{4}-x\right)+c$$

| cite | improve this answer | |
  • $\begingroup$ 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

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}

| cite | improve this answer | |
  • 1
    $\begingroup$ 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

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:

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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.