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).
- We don't need any explanation here; it is just right.
- 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.
- Since $\frac{1}{-2x}$ at $du$ is a constant, we can send it outside integral. We also integrate $\int u^{-1/2}=2\sqrt{u}$
- 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!
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 $x$!)