# Integration with respect to a function of x

You might think: what the hell are you talking about?

$\int&space;f(x)d\omega(x)$

Looks crazy, right? After this lesson it’s gonna look cool, trust me!

Let’s calculate the integral of x with respect to x².

$\int&space;x\,&space;dx^2$

One day I thought: “what is the integral of a function of x with respect to x²? That’s impossible I guess.” A couple of months later, that is less than a week ago, I realised: “substitution!”. If we know how to integrate with respect to x, u, t or whatever letter it is, then let’s let x² be equal to another letter, like α, (it can be any letter), and substitute the x inside the function with sqrt(α), since x²=α. Let me show you:

$\alpha&space;=&space;x^2\rightarrow&space;x=\sqrt\alpha$ $\int&space;\sqrt\alpha&space;\,&space;d\alpha=\int&space;\alpha^{\frac{1}{2}}\,&space;d\alpha=\frac{2}{3}\alpha^{\frac{3}{2}}\rightarrow&space;\frac{2}{3}\left&space;(&space;x^2&space;\right&space;)^{\frac{3}{2}}=\frac{2}{3}x^3+C$

That’s it! Let’s make things more interesting:

$\int&space;x\,&space;d\ln(x)$ $\mu&space;=\ln(x)\rightarrow&space;x=e^\mu$ $\int&space;e^\mu\,&space;d\mu=e^\mu\rightarrow&space;e^{\ln(x)}=x+C$
$\int&space;x&space;\,&space;d\sin(x)$ $\varphi=\sin(x)\rightarrow&space;x=\arcsin(\varphi)$ $\int&space;\arcsin(\varphi)d\varphi$

Let’s perform integration by parts:

$u=\arcsin(\varphi)\;\;\;\;\;dv&space;=&space;d\varphi$ $du=\frac{1}{\sqrt{1-\varphi^2}}d\varphi\;\;\;\;\;v&space;=&space;\varphi$ $\int&space;\frac{\varphi}{\sqrt{1-\varphi^2}}d\varphi$

Let’s perform a substitution:

$\tau=1-\varphi^2\;&space;\;&space;\;&space;\;&space;\;&space;d\tau=-2\varphi&space;d\varphi\rightarrow&space;d\varphi=-\frac{1}{2\varphi}d\tau$ $\int&space;\frac{\varphi}{\sqrt\tau}\left&space;(&space;-\frac{1}{2\varphi}&space;\right&space;)d\tau=-\int\frac{d\tau}{\sqrt&space;\tau}=-\int&space;\tau^{-\frac{1}{2}}d\tau=-2\tau^\frac{1}{2}$

Let’s undo the substitutions:

$-2\tau^\frac{1}{2}\rightarrow&space;-2\sqrt{&space;1-\varphi^2&space;}\rightarrow&space;-2\sqrt{&space;1-\sin^2(x)&space;}+C$

Not that hard, is it?

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