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