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As my question says: Where am I going wrong with this integral?

$$\int\frac{1}{\sqrt{1-x^2}}dx=\int(1-x^2)^{-1/2}dx=\int \frac{u^{-1/2}du}{-2x}=-\frac{1}{2x}2\sqrt{u}=-\frac{\sqrt{1-x^2}}{x}$$ For each step I will say explanation (which is wrong, but I can't find the solution).

  1. We don't need any explanation here; it is just right.
  2. Let's say, that $u=1-x^2$. Then $\frac{du}{dx}=\frac{d}{dx}(1-x^2)=-2x$. Then we insert this into our integral.
  3. Since $\frac{1}{-2x}$ at $du$ is a constant, we can send it outside integral. We also integrate $\int u^{-1/2}=2\sqrt{u}$
  4. Then we cancel out $2$ in denominator and numerator. We insert $u=1-x^2$.

So this is the end of the integral. But I know, that $\int\frac{1}{\sqrt{1-x^2}}dx=\arcsin{x}$. Where am I going wrong?

Of course, we can calculate the derivative of $\frac{d}{dx}(-\frac{\sqrt{1-x^2}}{x})$ to see, if we are right.

$$\frac{d}{dx}(-\frac{\sqrt{1-x^2}}{x})=-\frac{d}{dx}\sqrt{1-x^2}\times x - \sqrt{1-x^2}=-(\frac{1}{2\sqrt{1-x^2}}\times x\times (-2x)-\sqrt{1-x^2})=\frac{1}{\sqrt{1-x^2}}$$

But I get the same, so I am going wrong in both expressions. (I know, that $\arcsin{x}\neq \frac{1}{\sqrt{1-x^2}}$.)

Thank you for your effort!

P.S. I watched some videos about solving this integral using trigonometric substitutions and I can solve it, but I can't find the error in this integral. You don't have to tell me the real solution of this integral, just error. Thank you!

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    $\begingroup$ $x$ is not a constant, you have to substitute back $x = \sqrt{1-u}$ and integrate w.r.t $u$. $\endgroup$ – Ak. Mar 16 '20 at 13:04
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    $\begingroup$ @Ak19 Thank you! $\endgroup$ – User123 Mar 16 '20 at 13:06
  • $\begingroup$ By the way, when you tried to check your result using the derivative, you have forgotten the denominator of the quotient rule $\endgroup$ – Reinhard Meier Mar 16 '20 at 13:09
  • $\begingroup$ @ReinhardMeier Oh, I thought, that I need to calculate $(\frac{d}{dx}x)^2=1$. But I needed to calculate just $x^2$. Thank you! $\endgroup$ – User123 Mar 16 '20 at 17:21
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The error is when you say "Since $\frac{1}{-2x}$ is a constant at $du$, we can send it outside the integral." The fact is, $\frac{1}{-2x}$ is absolutely not constant with respect to $u$, since $u$ is a function of $x$.

(This would be even more clear if it were a definite integral. Instead of getting a number as you should, you'd somehow get a function of the dummy variable $x$!)

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