# Substitution

## Introduction

Recall that by chain rule the derivative of a composition of functions \( f(x)=u(v(x)) \) can be calculated as \( f^{\prime}(x)=u^{\prime}(v(x))\cdot v^{\prime}(x) \). Applying this to \( f(x)=\sin (x^2) \) we obtain \( f^{\prime}(x)=\cos (x^2)\cdot 2x \).

Assume now that we are asked to find \( \int 2x\cos (x^2)\,dx \). Well, we could just say ``notice that the antiderivative is \( \sin(x^2)+C \),’’ since the function \( \sin(x^2) \) is still fresh in our memory. We are now going to find a systematic way to treat the problems of this sort. More precisely, we will now learn how to find antiderivatives of functions like \( xe^{x^2} \), \( x^3e^{x^4} \), and many others.

## The method of substitution

** Example **

Find the integral \[ \int \cos \left(x^3\right)\cdot x^2\,dx.\]

Let us use the substitution \( y=x^3 \). Then we have \( x=y^{\frac13} \) and \( dx=\frac13\cdot y^{-\frac23}\,dy \).
Our next step is to transform the original integral into the one whose variable is \( y \). We will do this by replacing each occurrence of \( x \) with \( y^{\frac13} \). The integral now becomes \[ \int \cos (x^3)\cdot x^2\,dx=\int \cos y\cdot \left(y^{\frac13}\right)^2\cdot \frac13\cdot y^{-\frac23}\,dy=\frac13\int \cos y\,dy=\sin y+C.\]
We can now recover the antiderivative of \( \cos (x^3)\cdot x^2 \) by substituting back \( y=x^3 \) in the last expression. Therefore: \[ \int \cos \left(x^3\right)\cdot x^2\,dx=\sin y+C=\sin\left(x^2\right)+C.\]

In the end we provide the theorem responsible for substitution. It may look complicate on the first sight, but it is just formalizing what was done before. Its proof is a straight-forward application of the chain rule.

** Theorem **

Assume that \( F(x) \) is an antiderivative of the function \( f(x) \). If \( u \) is an invertible differentiable function such \( G(y) \) is an antiderivative of \( f(u(y))\cdot u^{\prime}(y) \), then \( F(x)=G\left(u^{-1}(x)\right)+C \).

It suffices to verify that the derivative of \( G\left(u^{-1}(x)\right)+C \) is equal to \( f(x) \). We use the chain rule to obtain:
\[ \frac{d}{dx} G\left(u^{-1}(x)\right)=G^{\prime}\left(u^{-1}(x)\right)\cdot \frac{d}{dx}u^{-1}(x)=f\left(u\left(u^{-1}(x)\right)\right)\cdot u^{\prime}\left(u^{-1}(x)\right)\cdot \frac{d}{dx}u^{-1}(x)=f(x)\cdot \frac{d}{dx}\left(u\left(u^{-1}(x)\right)\right)=f(x).\]