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I'm trying to find out the antiderivative. My approach is:

$\int \sqrt{\csc x-\sin x} dx = \int \sqrt{\frac{1}{\sin x}-\sin x} dx= \int \sqrt{\frac{1-\sin^2x}{\sin x}}dx $

Then: $\int \sqrt{\frac{\cos^2 x}{\sin x}} dx = \int \frac{\cos x}{\sqrt{\sin x}} dx $

Let $u$ be $\sin x$ so $\int \frac{du}{\sqrt{u}} = 2\sqrt{u} + k$

Finally $\int \sqrt{\csc x-\sin x} dx = 2\sqrt{\sin x} +k$

I thought I was right but apparently neither WolframAlpha nor the other antiderivative calculators agree with my result.

I don't know where I went wrong, I'd very much appreciate if someone could help me out.

Thanks BTW: I haven't mastered LaTeX yet, so forgive me if it is poorly formatted.

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    $\begingroup$ As long as we keep in mind that in fact $\;\sqrt{\cos^2x}=|\cos x|\;$ , your answer looks correct to me (up to a sign depending on the above) $\endgroup$ – DonAntonio Aug 29 '16 at 21:35
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    $\begingroup$ @Genis Remember that these online tools sometimes use different trig identities than we would and thus they may come up with an anti derivative that is essentially the same (I hope so!) but differ by a constant. $\endgroup$ – imranfat Aug 29 '16 at 21:39
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    $\begingroup$ FWIW, Maple agrees with your result. $\endgroup$ – user307169 Aug 29 '16 at 21:39
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    $\begingroup$ Your result is correct. $\endgroup$ – Mark Viola Aug 29 '16 at 21:40
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Your answer is correct and Wolfram Alpha does indeed agree with you. When I use Wolfram Alpha, I get this: $$ \int\sqrt{\csc x - \sin x} \, dx = 2\tan x \sqrt{\cos x \cot x} + C $$

Ignoring the $+C$ (and ignoring the $\sqrt{a^2} = |a|$ paradigm with trig functions, as we often do with indefinite integrals involving trig functions and substitutions, etc...) the RHS can be simplified as follows:

\begin{align} 2\tan x \sqrt{ \cos x \cot x} &= 2 \frac{\sin x}{\cos x} \sqrt{\cos x \frac{\cos x}{\sin x}}\\[0.3cm] &= 2\frac{\sin x}{\cos x}\sqrt{\frac{\cos^2 x}{\sin x}}\\[0.3cm] &= 2\frac{\sin x}{\cos x} \cdot \frac{\cos x}{\sqrt{\sin x}}\\[0.3cm] &= 2\sqrt{\sin x} \end{align}

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    $\begingroup$ Thanks, I did that too, but since WolframAlpha didn't simplify that I thought I was wrong, perhaps I should trust myself next time. Thanks for all. $\endgroup$ – Genis Aug 29 '16 at 21:50
  • $\begingroup$ You can also graph both functions (or their difference). Though this is not a proof, it can help you gain confidence in your answer if the graph is the 0 function (if you graph the difference) or two parallel functions. Then you may do the work @tilper has done; he could also have simplified the difference of the two answers. $\endgroup$ – Bernard Masse Aug 29 '16 at 21:56
  • $\begingroup$ Re: the above suggestion of simplifying difference of answers, it's very fun to have Maple do that. Have Maple evaluate your integral, see the answer appears to be different from yours. Then type "simplify(% - [insert your answer here]);", hit enter, see the 0, instant euphoria. $\endgroup$ – user307169 Aug 29 '16 at 22:24

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