# 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 $$F(x)=G\left(u^{-1}(x)\right)+C$$.

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