Caluclus: Table of contents



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


Determine 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 \[\int \cos (x^3)\cdot x^2\,dx=\int \cos y\cdot \left(y^{\frac13}\right)^2\cdot \frac13\cdot y^{-\frac23}\,dy=\frac13\int \cos y\,dy=\frac13\sin y+C.\] We can now recover the antiderivative of \(\cos (x^3)\cdot x^2\) by substituting back \(y=x^3\) in the last expression. Therefore: \[\int \cos \left(x^3\right)\cdot x^2\,dx=\frac13\sin y+C=\frac13\sin\left(x^3\right)+C.\]

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

Practice problems

Problem 1. Determine the indefinite integral \[\int \cos(5x)\,dx.\]

Problem 2. Determine the indefinite integral \[\int x\sqrt{1-x^2}\,dx.\]

Problem 3. Determine the indefinite integral \[\int x^4\sin(x^5)\,dx.\]

Problem 4. Determine the indefinite integral \[\int \frac{x}{x^2+1}\,dx.\]

Problem 5. Find the indefinite integral \[\int \frac{\sin\sqrt x}{\sqrt x}\,dx.\]