26. Integration by Substitution

Dated: 30-10-2024

Let's say we have

\[\frac{d}{du} g(u) = f(u)\]

Then

\[\int f(u) du = \int \left(\frac{d}{du}g(u)\right) du\]

\[= \int d \,g(u)\]

\[= g(u) + C\]

Now let us assume that \(u\) is a function¹ of \(x\). i.e. \(u(x)\). Then,

\[\frac{d}{dx} g(u) = \frac{d}{du} g(u) \cdot \frac{du}{dx} = f(u) \cdot \frac{du}{dx}\]

We can use this conclusion in a very specific scenario. Let us take a look

\[\int \left( f(u) \frac{du}{dx} \right) dx = \int \frac{d}{dx} (g(u)) dx = g(u) + C \]

Example

\[\int (x^2 + 1)^{50} \cdot 2x \cdot dx\]

Now let us define some variables.

So if we replace our data with the variables, we get

\[\int f(u) \cdot \frac{du}{dx} \cdot dx = \int f(u) du\]

This suggests we just need to integrate² \(f(u)\).

\[\int f(u) du = \int u^{50} du\]

\[= \frac{u^{50 + 1}}{50 + 1} + C\]

\[= \frac{u^{51}}{51} + C\]

Substituting \(u\), we get

\[= \frac{(x^2 + 1)^{51}}{51} + C\]