Substitution Magic

Big Idea

The substitution $u = g(x)$ allows us to simplify integrals:

$\displaystyle \int f'(g(x)) \cdot g'(x)\, \mathrm dx = \int f(u)\, \mathrm du.$

In particular, we recover two useful integration forms:

$\displaystyle \int f'(ax+b)\, \mathrm dx = \frac 1a \cdot f'(ax+b) +C,\quad \int \frac{f'(x)}{f(x)}\, \mathrm dx = \ln|f(x)| +C.$

Questions

Question 1. Using $u = -x^2$ , evaluate the integral $\displaystyle \int xe^{-x^2}\, \mathrm dx$ .

Deduce the value of $\displaystyle \int_0^1 xe^{-x^2}\, \mathrm dx$ .

(Click for Solution)

Solution. Making the substitution $u = -x^2 \Rightarrow -\frac 12 \mathrm du = x\, \mathrm dx$ ,

$\begin{aligned} \int xe^{-x^2}\, \mathrm dx &= \int e^{-x^2} \cdot x\, \mathrm dx\\ &= \int e^u\left( -\frac 12 \right) \, \mathrm du \\ &= -\frac 12 \int e^u\, \mathrm du \\ &= -\frac 12 \cdot e^u + C \\ &= -\frac 12 \cdot e^{-x^2} + C. \end{aligned}$

Hence,

$\begin{aligned} \int_0^1 xe^{-x^2}\, \mathrm dx &= \left[ -\frac 12 \cdot e^{-x^2} \right]_0^1 \\ &= \left( -\frac 12 \cdot e^{-1^2} \right) - \left( -\frac 12 \cdot e^{-0^2} \right) \\ &= \frac 12(1-e^{-1}). \end{aligned}$

Question 2. Evaluate the integral

$\displaystyle \int \sin(\cos^2(x)) \cos(1 - \sin^2(x)) \sin(2x)\, \mathrm dx.$

(Click for Solution)

Solution. Make the substitution

$u = \cos^2(x) = 1-\sin^2(x) = \frac 12(1 + \cos 2x)$

so that

$\begin{aligned} \frac{ \mathrm du }{\mathrm dx} &= \frac 12 (0 + (-\sin(2x)) \cdot 2) \\ &= - \sin (2x) \\ \quad (-1)\, \mathrm du &= \sin(2x)\, \mathrm dx. \end{aligned}$

Hence,

$\begin{aligned} &\int \sin(\cos^2(x)) \cos(1 - \sin^2(x)) \sin(2x)\, \mathrm dx \\ &= \int \sin(\cos^2(x)) \cos(\cos^2(x)) \sin(2x)\, \mathrm dx\\ &= \int \sin(u) \cos(u) \cdot (-1)\, \mathrm du \\ &= -\frac 12 \int 2 \sin(u) \cos(u) \, \mathrm du \\ &= -\frac 12 \int \sin(2u) \, \mathrm du \\ &= -\frac 12 \cdot \frac 12 \cdot (-\cos(2u)) + C \\ &= \frac 14 \cdot \cos(2u) + C \\ &= \frac 14 \cdot \cos(2 \cos^2(x)) + C.\end{aligned}$

Question 3. Evaluate the integral $\displaystyle \int x^x(\ln x + 1)\, \mathrm dx$ .

(Click for Solution)

Solution. Using the laws of exponents $a^b = e^{b \ln a}$ ,

$\displaystyle \int x^x(\ln x + 1)\, \mathrm dx = \int e^{x \ln x }(\ln x + 1)\, \mathrm dx.$

Then make the substitution $u = x \ln x$ so that

$\begin{aligned} \frac{\mathrm d u}{\mathrm dx} &= \frac{\mathrm d}{\mathrm dx} (x \ln x) \\ &= \ln x +1 \\ \mathrm du &= (\ln x + 1)\, \mathrm dx.\end{aligned}$

Since $e^u = e^{x \ln x} = x^x$ ,

$\displaystyle \int x^x(\ln x + 1)\, \mathrm dx = \int e^u\, \mathrm du = e^u + C = x^x + C.$

—Joel Kindiak, 4 Sept 25, 1644H

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