5.19 Further Integration by SubstitutionHL ONLY

1. Standard Substitution Recap

We have seen the simple case of substitution $u=f(x)$, when the derivative of $f(x)$ is part of the integrand.

For example, let $u = f(x)$. Then $\dfrac{du}{dx} = f'(x) \implies du = f'(x)dx$.

$$ \begin{aligned} I &= \int \dfrac{f'(x)}{f(x)} dx = \int \dfrac{1}{u} du = \ln(u) + c \\[0.2cm] &= \ln(f(x)) + c \end{aligned} $$

In this case, we can omit the whole process and give the result directly.

Example 1

Find $A = \displaystyle\int \dfrac{\cos x}{\sin x + 1} dx$ and $B = \displaystyle\int \dfrac{x}{x^2 + 1} dx$.

Since the derivative of $u = \sin x + 1$ is $\cos x$ $$ A = \int \dfrac{\cos x}{\sin x + 1} dx = \ln(\sin x + 1) + c $$ For $B$, the derivative of $u = x^2 + 1$ is $2x$, so $$ B = \int \dfrac{x}{x^2 + 1} dx = \dfrac{1}{2} \int \dfrac{2x}{x^2 + 1} dx = \dfrac{1}{2} \ln(x^2 + 1) + c $$

Similarly, we can use direct substitution for other standard forms:

$\displaystyle\int f'(x) e^{f(x)} dx = e^{f(x)} + c$
$\displaystyle\int f'(x) \sin(f(x)) dx = -\cos(f(x)) + c$
$\displaystyle\int f'(x) \cos(f(x)) dx = \sin(f(x)) + c$
$\displaystyle\int f'(x) (f(x))^n dx = \dfrac{(f(x))^{n+1}}{n+1} + c$

Example 2

Find the following integrals using direct substitution methods.

$\displaystyle\int \cos x e^{\sin x} dx = e^{\sin x} + c$

$\displaystyle\int 3x^2 e^{x^3 + 2} dx = e^{x^3 + 2} + c$

$\displaystyle\int x^2 e^{x^3 + 2} dx = \dfrac{1}{3} \int 3x^2 e^{x^3 + 2} dx = \dfrac{1}{3} e^{x^3 + 2} + c$

$\displaystyle\int 2x \sin(x^2 + 3) dx = -\cos(x^2 + 3) + c$

$\displaystyle\int 5x \sin(x^2 + 3) dx = \dfrac{5}{2} \int 2x \sin(x^2 + 3) dx = -\dfrac{5}{2} \cos(x^2 + 3) + c$

$\displaystyle\int 2x(x^2 + 3)^5 dx = \dfrac{(x^2 + 3)^6}{6} + c = \dfrac{1}{6} (x^2 + 3)^6 + c$

$\displaystyle\int x(x^2 + 3)^5 dx = \dfrac{1}{2} \int 2x(x^2 + 3)^5 dx = \dfrac{1}{2} \dfrac{(x^2 + 3)^6}{6} + c$

$\displaystyle\int 2x \sqrt{x^2 + 3} dx = \int 2x(x^2 + 3)^{1/2} dx = \dfrac{(x^2 + 3)^{3/2}}{3/2} + c$

$\displaystyle\int \dfrac{3(\ln x)^2}{x} dx = \int \dfrac{1}{x} 3(\ln x)^2 dx = (\ln x)^3 + c$

$\displaystyle\int \dfrac{(\ln x)^3}{2x} dx = \dfrac{1}{2} \int \dfrac{1}{x} (\ln x)^3 dx = \dfrac{1}{2} \dfrac{(\ln x)^4}{4} + c = \dfrac{1}{8} (\ln x)^4 + c$

2. Reusing $u=f(x)$ to Isolate $x$

We will see two more cases of substitution. Case 1 involves reusing $u=f(x)$ to get rid of all remaining $x$ terms in the integrand. Characteristic examples of this case are integrals of the form $$ \int \text{polynomial} \cdot \sqrt{ax+b} \, dx \quad \text{or} \quad \int \dfrac{\text{polynomial}}{(ax+b)^n} \, dx $$ where we let $u = ax + b$.

Example 3

Find $I = \displaystyle\int x^3 \sqrt{x^2 + 3} dx$.

Use the substitution $u = x^2 + 3$. Then $$ \dfrac{du}{dx} = 2x \implies dx = \dfrac{du}{2x} $$ Thus, $$ I = \int x^3 \sqrt{u} \dfrac{du}{2x} = \dfrac{1}{2} \int x^2 \sqrt{u} du $$ The result still contains $x^2$, but $u = x^2 + 3 \implies x^2 = u - 3$. Thus, $$ \begin{aligned} I &= \dfrac{1}{2} \int (u - 3) \sqrt{u} du = \dfrac{1}{2} \int (u - 3) u^{1/2} du = \dfrac{1}{2} \int (u^{3/2} - 3u^{1/2}) du \\ &= \dfrac{1}{2} \left( \dfrac{2}{5} u^{5/2} - 3 \dfrac{2}{3} u^{3/2} \right) + c = \dfrac{1}{5} u^{5/2} - u^{3/2} + c \\ &= \dfrac{1}{5} (x^2 + 3)^{5/2} - (x^2 + 3)^{3/2} + c \end{aligned} $$

Example 4

Find $I = \displaystyle\int \dfrac{x^2}{x + 2} dx$.

Let $u = x + 2$ (so that $x = u - 2$). Then $$ \dfrac{du}{dx} = 1 \implies dx = du $$ Thus, $$ \begin{aligned} I &= \int \dfrac{(u - 2)^2}{u} du = \int \dfrac{u^2 - 4u + 4}{u} du = \int \left( u - 4 + \dfrac{4}{u} \right) du \\ &= \dfrac{u^2}{2} - 4u + 4 \ln u + c = \dfrac{(x + 2)^2}{2} - 4(x + 2) + 4 \ln(x + 2) + c \end{aligned} $$

Example 5

Another popular substitution of this kind is $u = e^x$ due to its simple derivative. Find $$ I_1 = \int \dfrac{e^x}{e^{2x} + 4} dx \quad \text{and} \quad I_2 = \int \dfrac{e^{2x}}{e^x + 4} dx $$

For both integrals, let $u = e^x$, then $$ \dfrac{du}{dx} = e^x \implies dx = \dfrac{du}{e^x} $$ For $I_1$, $$ \begin{aligned} I_1 &= \int \dfrac{e^x}{u^2 + 4} \dfrac{du}{e^x} = \int \dfrac{1}{u^2 + 4} du = \dfrac{1}{2} \arctan\left(\dfrac{u}{2}\right) + c \\ &= \dfrac{1}{2} \arctan\left(\dfrac{e^x}{2}\right) + c \end{aligned} $$ For $I_2$, $$ \begin{aligned} I_2 &= \int \dfrac{u^2}{u + 4} \dfrac{du}{u} = \int \dfrac{u}{u + 4} du = \int \dfrac{u + 4 - 4}{u + 4} du = \int \left( 1 - \dfrac{4}{u + 4} \right) du \\ &= u - 4 \ln|u + 4| + c = e^x - 4 \ln(e^x + 4) + c \end{aligned} $$

3. Inverse Substitution $x = g(u)$

Case 2 involves setting $x = \text{expression of } u$ instead of $u = \text{expression of } x$. In this case, the substitution is not obvious and is usually given in an exam. We will see characteristic substitutions of this kind.

Example 6

For $I = \displaystyle\int \dfrac{1}{x^2 + 4} dx$, use the substitution $x = 2 \tan u$.

We have $$ \dfrac{dx}{du} = 2 \sec^2 u \implies dx = 2 \sec^2 u \, du $$ Thus, $$ \begin{aligned} I &= \int \dfrac{1}{(2 \tan u)^2 + 4} (2 \sec^2 u \, du) = \int \dfrac{2 \sec^2 u}{4 \tan^2 u + 4} du \\ &= \int \dfrac{2 \sec^2 u}{4(\tan^2 u + 1)} du = \int \dfrac{2 \sec^2 u}{4 \sec^2 u} du \\ &= \dfrac{1}{2} u + c = \dfrac{1}{2} \arctan\left(\dfrac{x}{2}\right) + c \end{aligned} $$

Example 7

For $I = \displaystyle\int \dfrac{1}{\sqrt{4 - x^2}} dx$, use the substitution $x = 2 \sin u$.

We have $$ \dfrac{dx}{du} = 2 \cos u \implies dx = 2 \cos u \, du $$ Thus, $$ \begin{aligned} I &= \int \dfrac{1}{\sqrt{4 - (2 \sin u)^2}} (2 \cos u \, du) = \int \dfrac{2 \cos u}{\sqrt{4 - 4 \sin^2 u}} du \\ &= \int \dfrac{2 \cos u}{\sqrt{4(1 - \sin^2 u)}} du = \int \dfrac{2 \cos u}{\sqrt{4 \cos^2 u}} du \\ &= \int \dfrac{2 \cos u}{2 \cos u} du = u + c = \arcsin\left(\dfrac{x}{2}\right) + c \end{aligned} $$

In general, the two formulas from the formula booklet can be shown by using the substitutions $x = a \tan u$ and $x = a \sin u$ respectively.

$\displaystyle\int \dfrac{1}{x^2 + a^2} dx = \dfrac{1}{a} \arctan\left(\dfrac{x}{a}\right) + c$
$\displaystyle\int \dfrac{1}{\sqrt{a^2 - x^2}} dx = \arcsin\left(\dfrac{x}{a}\right) + c$

We can see these two substitutions in similar cases. See expression $\sqrt{a^2 + x^2}$, use substitution $x = a \tan \theta$. See expression $\sqrt{a^2 - x^2}$, use substitution $x = a \sin \theta$.

Example 8

Find $I = \displaystyle\int \sqrt{16 - x^2} dx$, by using the substitution $x = 4 \sin \theta$.

We have $$ \dfrac{dx}{d\theta} = 4 \cos \theta \implies dx = 4 \cos \theta \, d\theta $$ Thus, $$ \begin{aligned} I &= \int \sqrt{16 - (4 \sin \theta)^2} (4 \cos \theta \, d\theta) \\ &= \int \sqrt{16(1 - \sin^2 \theta)} (4 \cos \theta) d\theta \\ &= \int (4 \cos \theta)(4 \cos \theta) d\theta \\ &= \int 16 \cos^2 \theta d\theta \end{aligned} $$ We use the double angle identity $\cos^2 \theta = \dfrac{1 + \cos 2\theta}{2}$. Thus:
$$ \begin{aligned} I &= \int 16 \left( \dfrac{1 + \cos 2\theta}{2} \right) d\theta = 8 \int (1 + \cos 2\theta) d\theta \\ &= 8 \left( \theta + \dfrac{\sin 2\theta}{2} \right) + c = 8\theta + 4 \sin 2\theta + c \end{aligned} $$ But $\theta = \arcsin\left(\dfrac{x}{4}\right)$ and $$ 4 \sin 2\theta = 8 \sin \theta \cos \theta = 8 \left(\dfrac{x}{4}\right) \cos \theta = 2x \cos \theta $$ Since $\sin \theta = \dfrac{x}{4}$, we construct a right triangle. The adjacent side is $\sqrt{16 - x^2}$, so $$ \cos \theta = \dfrac{\sqrt{16 - x^2}}{4} \implies 2x \cos \theta = 2x \left(\dfrac{\sqrt{16 - x^2}}{4}\right) = \dfrac{x \sqrt{16 - x^2}}{2} $$ Therefore, $$ I = 8 \arcsin\left(\dfrac{x}{4}\right) + \dfrac{x \sqrt{16 - x^2}}{2} + c $$