Reading Courant's Calculus textbook, he mentions a type of substitution to solve an integral

$$\int_a^b h(\phi(u))du.$$

If $\phi'(u) \neq 0$ on $[a,b]$ then there is a function $\psi(x)$ inverse to $\phi$, and we can write $f(x) = h(x)\psi'(x)$, then

$$\int h(\phi(u))du = \int \frac{f(\phi(u))}{\psi'(x)} du = \int f(\phi(u))\phi'(u) du = \int f(x) dx.$$

He gives as an example of this

$$\int_1^4 \frac1{\sqrt{x}}dx \overset{(u^2=x)}{=}\int_1^4\frac1{u}dx = \int_1^2\frac{2u}{u}du = 2.$$

Also he gives the example of

$$\int \sin{2x}.$$

My Question: These seem like kind of trivial examples. Can someone provide some examples of instances (or provide a reference) where such substitutions are helpful? Thanks

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    $\begingroup$ this is the very basis of a change of variables; it is equivalent to $u$-substitution only it makes you more conscious of the invertibility requirement $\endgroup$ – oldrinb Aug 26 '13 at 23:09

If I'm reading this correctly, this is just another way of writing a "reverse substitution", so to speak, where instead of substituting $u$ as an expression in $x$, you replace $x$ by writing $x$ as an expression in $u$. One common example from one-variable calculus is the trigonometric substitution. When evaluating $$ \int_0^1 \sqrt{1 - x^2} \, dx$$ you can perform the reverse substitution $x = \sin{u}, dx = \cos{u} \, du$ to make the integral into a standard trigonometric integral.


What about $\int\frac{x^{2}-1}{x^{2}+1}\frac{1}{\sqrt{x^{4}+1}}$? I think this provides an example of a non-trivial integration problem which uses reverse substitution in its solution. (This integral is not completely solved by using reverse substitution but it brings it into a reasonable form through this method)

Take $x=\sqrt{\frac{1+u}{1-u}}$ then $dx=\sqrt{\frac{1-u}{1+u}}\frac{1}{(1-u)^{2}}du$. Putting this into the integral we get:

$\int\frac{(\frac{1+u}{1-u}-1)}{\frac{1+u}{1-u}+1}\frac{1}{\sqrt{1+\frac{(1+u)^{2}}{(1-u)^{2}}}}\sqrt{\frac{1-u}{1+u}}\frac{1}{(1-u)^{2}}du=\frac{1}{\sqrt{2}}\int u\frac{1-u}{\sqrt{1+u^{2}}}\sqrt{\frac{1-u}{1+u}}\frac{1}{(1-u)^{2}}du$

$=\frac{1}{\sqrt{2}}\int u\frac{1}{\sqrt{1+u^{2}}}\frac{1}{\sqrt{1-u^{2}}}$

Take $w=u^{2}$ then the above becomes:

$=\frac{1}{2\sqrt{2}}\int\frac{1}{\sqrt{1-w^{2}}}dw$ where the last integral is solvable by trigonometric substitution.


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