Integrating $\int\frac{1}{\sqrt{1-x^2}}\,dx$ using $u$-substitution with an $x$ left over after?

Question

$$\int\frac{1}{\sqrt{1-x^2}}\,dx.$$

When I use $u$-substitution to get the indefinite integral of that function, I get an extra $x$ on my $du$. How can I deal with that $x$?

This is what I got: $\frac{1}{2}du = x \, dx$

Answer 1

$u$-substitution could work, but as Tim said it is much easier to just recognize that $\frac{1}{\sqrt{1-x^2}}$ is the derivative of an inverse trig function.

If you insist on using $u$-sub, you can always write $x=\sin\theta$, so that $dx=\cos \theta \,d\theta$. Then, $\sqrt{1-x^2} = \sqrt{1-\sin^2\theta} = \cos\theta$, so that our problem is just: $$\int\frac{dx}{\cos\theta}=\int\frac{\cos\theta \, d\theta}{\cos\theta} = \int1 \, d\theta = \theta + C = \arcsin(x) + C$$

Answer 2

$u$ substitution will not work for this problem. You need to recognize that $1/\sqrt{1-x^2}$ is the derivative of an inverse trig function.