# Where am I going wrong with the integral $\int\frac{1}{\sqrt{1-x^2}}dx$?

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 $$\arcsin{x}\neq \frac{1}{\sqrt{1-x^2}}$$.)

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!

• $x$ is not a constant, you have to substitute back $x = \sqrt{1-u}$ and integrate w.r.t $u$. – Ak. Mar 16 '20 at 13:04
• @Ak19 Thank you! – User123 Mar 16 '20 at 13:06
• By the way, when you tried to check your result using the derivative, you have forgotten the denominator of the quotient rule – Reinhard Meier Mar 16 '20 at 13:09
• @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! – User123 Mar 16 '20 at 17:21

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 $$x$$!)