Theory

Theory 1

The method of $u$-substitution is applicable when the integrand is a product, with one factor a composite whose inner function’s derivative is the other factor.

Substitution

Suppose the integral has this format, for some functions $f$ and $u$:

$$\int f(u(x))\cdot u^{\prime }(x)dx$$

Then the rule says we may convert the integral into terms of $u$ considered as a variable, like this:

$$\int f(u(x))\cdot u^{\prime }(x)dx\gg \gg \int f(u)du$$

The technique of $u$-substitution comes from the chain rule for derivatives:

$$\frac{d}{dx}F(u(x))=f(u(x))\cdot u^{\prime }(x)$$

Here we let $F^{\prime }=f$. Thus ${\displaystyle \int f(x)dx=F(x)+C}$ for some $C$.

Now, if we integrate both sides of this equation, we find:

$$F(u(x))=\int f(u(x))\cdot u^{\prime }(x)dx$$

And of course $F\left(u\right)={\displaystyle \int f(u)du-C}$.

Extra - Full explanation of $u$-substitution

(1) Chain rule for derivatives:

Let $F(x)$ be a function and $F^{\prime }=f$ its derivative. Let $u(x)$ be another function.

Using primes:

$$(F\circ u{)}^{\prime }=(F^{\prime }\circ u)\cdot u^{\prime }$$

Using differentials:

$$\frac{d}{dx}F(u(x))=f(u(x))\cdot u^{\prime }(x)$$

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(2) Integrate both sides:

$$\begin{array}{cc}\frac{d}{dx}F(u(x))=f(u(x))\cdot u^{\prime }(x)& {\gg \gg }^{\int }\int \frac{d}{dx}F(u(x))=\int f(u(x))\cdot u^{\prime }(x)\\ & \\ & {\gg \gg }^{\text{FTC}}F(u(x))=\int f(u(x))\cdot u^{\prime }(x)\end{array}$$

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(3) Introduce ‘variable’ $u$ from the $u$-format of the integral:

Treating $u$ as a variable, the definition of $F$ gives:

$$F(u)=\int f(u)du+C$$

Set the ‘variable’ $u$ to the ‘function’ $u$ output:

$$F(u){|}_{u=u(x)}=F(u(x))$$

Combine these:

$$\begin{array}{cc}F(u(x))& =F(u){|}_{u=u(x)}\\ & \\ & ={\int f(u)du|}_{u=u(x)}+C\end{array}$$

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(4) Substitute for $F(u(x))$:

$$\int f(u(x))\cdot u^{\prime }(x)dx=F(u(x))={\int f(u)du|}_{u=u(x)}+C$$

This is “$u$-substitution” in final form.