Integrate $\frac{x}{\sqrt{1-x^2}}$

Question

I want to integrate

\begin{equation*} \int_{a}^b\frac{x}{\sqrt{1-x^2}}dx \end{equation*}

by using the substition $\varphi(t)=\sqrt{1-x^2}$. For sake of simplicity we assume $0< a< b< 1$. If we consider the constant function $f(x)=1$, then we get

\begin{equation*} \int_a^b\frac{x}{\sqrt{1-x^2}}dx=\int_{a}^{b}(-1)\cdot f(\varphi(t))\varphi(t)'dt=\int_{\varphi(a)^{-1}}^{\varphi(b)^{-1}}(-1)\cdot 1dt=\sqrt{1-a^2}-\sqrt{1-b^2}. \end{equation*} Is this correct? It feels strange to use $f(x)=1$ but I don't see that it violates any conditions, don't it?

Answer 1

There is no rule saying you cannot use $f(x)=1$. The only thing missing in the proof above is to show how you know that

$$\frac{x}{\sqrt{1-x^2}} = -f(\varphi(t))\varphi'(t)$$

is in fact true.

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Then again, if you found a function, $F$, such that $F'=f$, then you don't really need to use a "substitution" method if you want to calculate $$\int_a^b f(x)dx$$ since the fundamental theorem of calculus already tells you that the above value is just $F(b)-F(a)$.

Answer 2

Your solution is correct. The same phenomenon will happen any time that the variable $\phi$ you choose for your substitution happens to be the antiderivative of the integrand.

Alternatively, substituting $$u = 1 - x^2, \qquad du = - 2x \,dx$$ transforms the integral to $$-\frac{1}{2} \int_{1 - a^2}^{1 - b^2} \frac{du}{\sqrt u} = -\left.\sqrt{u}\right\vert_{1 - a^2}^{1 - b^2} .$$

Answer 3

Let's first consider the integral without limits:

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

Now substitute:

$$ \phi(x) = x^2. $$

Then it holds:

$$ \int \frac{x}{\sqrt{1-x^2}} \, dx = \int \frac{1}{2\sqrt{1-\phi(x)}} \, d\phi(x). $$

The second integral can be integrated normally to:

$$ = -\sqrt{1-\phi(x)} + C. $$

Now substitute back to:

$$ = -\sqrt{1-x^2} + C. $$

It is therefore valid:

$$ \int_a^b \frac{x}{\sqrt{1-x^2}} \, dx = \left[-\sqrt{1-x^2}\right]_a^b, $$

and therefore:

$$ = -\sqrt{1-b^2} + \sqrt{1-a^2}. $$

This is valid for $0 < a < b < 1$.