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The task goes as following:

$∫xdx/\sqrt{1-x^2}$

Using substitution $t=1-x^2$:
$-1/2∫t^{-1/2}dt=-\sqrt t=-\sqrt{1-x^2}+C $

But when using substitute $x^2$ you get:
enter image description here

Why are the two solutions different? I can't find any mistakes.



Same for this one: $∫dx/(b^2x^2-a^2)$

When pulling out $b^2$ from under the root, and then out of the integral and using substitution
$t-x=\sqrt{x^2-(a/b)^2}$
$x=[t^2+(a/b)^2]/2t$
$dx=[t^2-(a/b)^2]/2t^2*dt$


I get $1/b*ln(x+\sqrt{x^2-(a/b)^2})+C$ as a result. But when introducing $bx=t$ as a substitute (like they did in my book) they got $1/b*ln(bx+\sqrt{b^2x^2-a^2})+C$. How did the entire argument of $ln$ get multiplied by b?

Thank you in advance.

EDIT: I figured you can't treat $t$ as $\sqrt{t^2}$ and simply integrate that as $arcsin$. The argument $x^2$ in $∫dx/\sqrt{1-x^2}$ must be "clean". I tried substituting $\sqrt{t}$ with $u$, but I ended up at the very beginning.

enter image description here


Clearly with substitute $t=x^2$ this task cannot be solved. Does anyone know what's the problem with the second one?

0lt
  • 657

1 Answers1

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The question regarding the first integral is solved above in the "EDIT" section. For the second integral I learned the answer today.

It turns out, when you single out $b$ from argument of $ln$, what you get is logarithm of a product $ln(b*...)$ which equals to $lnb+ln...+C$. Since $lnb$ is a constant, it's already been included in $C$.

Thanks everyone for your contribution.

0lt
  • 657