integration by substitution

To put this into practice is to integrate \[x^{3}\sqrt{x^{4}+16}\].

To solve \[\int x^{3}\sqrt{x^{4}+16}\,dx\], we first try to turn whatever that's inside, which is \[x^{3}\sqrt{x^{4}+16}\] into the form of \[f(g(x))\cdot g^{\prime}(x)\]. We define \[f(u)=\sqrt{u}\] (\[u\] is a placeholder value, has nothing to do with the \[u\]-subtitution we're doing later) and \[g(x)=x^{4}+16\].

Notice that \[g^{\prime}(x)=4x^{3}\], which makes \[f(g(x))\cdot g^{\prime}(x)=\sqrt{x^{4}+16}\cdot 4x^{3}\]. To fix the fact that we're gotten \[4x^{3}\] not \[x^{3}\], we just write that \[\int x^{3}\sqrt{x^{4}+16}\,dx=\frac{1}{4}\int 4x^{3}\sqrt{x^{4}+16}\,dx\]. The reason in which this is possible has been proven here.

Now that we've turned it into this form, we can now substitute everything we have into this equation \[\int f(g(x))\cdot g^{\prime}(x)\,dx=F(g(x))+C\]. Now, since \[F^{\prime}(x)=f(x)\], reversing that, \[f(u)=\sqrt{u}=F^{\prime}(u)\], making \[F(u)=\frac{u^{\frac{3}{2}}}{\frac{3}{2}}=\frac{2}{3}u^{\frac{3}{2}}\]. Substituting the rest, we get that \[\int \sqrt{x^{4}+16}\cdot 4x^{3}\,dx=\frac{2}{3}(x^{4}+16)^{\frac{3}{2}}+C\]. Accounting for the \[\frac{1}{4}\], we get our final answer as:

\begin{align*} \int \sqrt{x^{4}+16}\cdot x^{3}\,dx&=\frac{1}{4}\int \sqrt{x^{4}+16}\cdot 4x^{3}\,dx\\ &=\frac{1}{4}\cdot \frac{2}{3}\left( x^{4}+16 \right)^{\frac{3}{2}}+C\\ &=\frac{1}{6}\left( x^{4}+16 \right)^{\frac{3}{2}}+C\\ \end{align*}

This process is quite tedious, thus arises the \[u\]-substitution method as another way to implement the theorem above. The reason it works is that let \[u=g(x)\], then \[\int f(u)\cdot u^{\prime}\,dx=\int f(u)\cdot \frac{du}{dx}\,dx=\int f(u)\,du\].

Similarly, we substitute the most complicated part of the equation, \[u=x^{4}+16\] (\[u\] here is the equivalent of \[g(x)\]). \[u^{\prime}=\frac{du}{dx}=4x^{3}\], which implies that \[du=\frac{du}{dx}\cdot dx=4x^{3}\cdot dx\]. Then, \[\int x^{3}\sqrt{x^{4}+16}\,dx=\int \sqrt{x^{4}+16}\left( x^{3}\cdot dx \right)=\frac{1}{4}\int \sqrt{x^{4}+16}\left( 4x^{3}\cdot dx \right)\] (yes they represent the same thing). Substituting \[u\] and \[du\], we get \[\frac{1}{4}\int\sqrt{u}\,du=\frac{1}{4}\cdot \frac{2}{3}u^{\frac{3}{2}}=\frac{1}{6}\left( x^{4}+16 \right)^{\frac{3}{2}}+C\].

Note that the reason this convoluted notation works has something to do with differential forms.

Both of these steps are precisely the statement "I have \[\int f(g(x))\cdot g^{\prime}(x)\,dx\] therefore it must be \[F(g(x))+C\]".

Evaluate the integral \[\int_{0}^{1}e^{-2x}\,dx\].

As usual, we will use the more rigorous method to solve it first. Define \[f(u)=e^{u}\], \[g(x)=-2x\], then \[g^{\prime}(x)=-2\]. We also note that \[g(0)=0\] and \[g(1)=-2\]. Substituting what we have just defined, \[f(g(x))\cdot g^{\prime}(x)=e^{-2x}\cdot -2\]. Similarly, to fix the fact that we have an extra \[-2\], we simply write \[-\frac{1}{2}\int_{0}^{1} e^{-2x}\cdot -2\,dx\].

Now that that's done, based on what we've proven above,

\begin{align*} \int_{0}^{1}e^{-2x}\,dx&=-\frac{1}{2}\int_{0}^{1}e^{-2x}\cdot -2\,dx\\ &=-\frac{1}{2}(F(g(1))-F(g(0)))\\ \end{align*}

Since \[f(u)=e^{u}\], it follows that \[F(u)=e^{u}+C\], then,

\begin{align*} -\frac{1}{2}(F(g(1))-F(g(0)))&=-\frac{1}{2}\left( \left( e^{-2}+C \right)-\left( e^{0}+C \right) \right)\\ &=-\frac{1}{2}\left( e^{-2}-e^{0} \right)\\ &=\frac{1}{2}\left( 1-e^{-2} \right)\\ \end{align*}

Again, this entire process can be simplified with the \[u\]-substitution. First, pick, \[f(u)=e^{u}\], \[u=-2x\], then \[u^{\prime}=\frac{du}{dx}=-2\implies du=-2\cdot dx\]. We note that \[u(0)=0\] and \[u(1)=-2\]. Then we carry out the integration,

\begin{align*} \int_{x=0}^{x=1}e^{-2x}\,dx&=\int_{u=g(a)}^{u=g(b)}f(u)\,du\\ &=\int_{u=0}^{u=-2}e^{u}(-2\cdot\,dx)\\ &=\frac{1}{2}\int_{u=-2}^{u=0}e^{u}\cdot 1\,du\\ &=\frac{1}{2}\biggl[ e^{u} \biggr]_{u=-2}^{u=0}\\ &=\frac{1}{2}\left( 1-e^{-2} \right)\\ \end{align*}

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