Integration By Substitution

Integrating a complex function can often be done by substituting a simpler function and it’s derivative, known as derivative substitution (sometimes called u-substitution) or change of variable. It can be thought as of a counterpart (of sorts) to the Chain Rule.

Definition §

Assume we have two differentiable functions, $g$ and $f$. Then these two integrals are equivalent:

$${\int }_a^{b}f(g(x))g^{\prime }(x)dx={\int }_{g(a)}^{g(b)}f(u)du$$

Note the change of variables, the left hand side is an integral with respect to $u$ instead of $x$.

Why? If we substitute $u=g(x)$, and it follows that $\frac{du}{dx}=g^{\prime }(x)$, then we can simplify the integral:

$${\int }_a^{b}f(u)\times \frac{du}{dx}dx={\int }_{g(a)}^{g(b)}f(u)du$$

In other words, if we have a product of a function and it’s derivative, we can simplify the entire integral. This also applies to indefinite integrals

#todo WHY DOES THIS WORK?!

Examples §

1: Sin(x) with polynomials

Solve the following integral: $\int 4x^3\sin (x^4)dx$

We can rewrite the question as $\int f(g(x))g^{\prime }(x)dx$, where $f(x)=\sin (x)$ and $g(x)=x^4$ Let $u=g(x)=x^4$. Then $\frac{du}{dx}=4x^3$ Using the substitution method:

$\int 4x^3\sin (x^4)dx=\int \sin (u)\times \frac{du}{dx}dx$ $=\int \sin (u)du$ $=-\cos (u)+c$ $=-\cos (x^4)+c$

2: Definite Integral

Solve the following integral: ${\int }_0^{\frac{\pi }{2}}\sin (x)\cos (x)dx$
We can begin by rewriting the integral in the form $\int f(g(x))g^{\prime }(x)dx$. Here $g(x)=\sin (x)$ and $f(x)=x$.

Since it is known that $\frac{d}{dx}(\sin (x))=\cos (x)$, we can let $u=g(x)=\sin (x)$, knowing that $u^{\prime }=\frac{du}{dx}=g^{\prime }(x)=\cos (x)$. Hence the integral can be expressed in the form: ${\int }_0^{\pi /2}u\times \frac{du}{dx}dx$

We need to ensure the bounds of the integral, $a$ and $b$ are replaced with $g(a)$ and $g(b)$. $\sin (0)=0$ and $\sin \left(\frac{\pi }{2}\right)=1$

So our final integral is: ${\int }_0^{1}udu$ $={\left[\frac{u^2}{2}\right]}_0^1=\frac{1}{2}-0$ $=\frac{1}{2}$