5.10 Integration by Substitution

1. The Method of Substitution

Integration by substitution (called $u$-substitution) is a technique used to evaluate integrals of composite functions. It is the reverse process of the Chain Rule for differentiation. We use this method when an integrand contains a function alongside its derivative.

The Substitution Formula:

$$ \int f(g(x))g'(x) dx = \int f(u) du $$
METHODOLOGY:
The goal is to replace $x$ and $dx$ from the original integral and obtain a simpler integral in terms of $u$ and $du$.
  • Choose a substitution $u=g(x)$. The criterion is that $g'(x)$ must exist in the integral as a factor.
  • Find the derivative: $\dfrac{du}{dx} = g'(x) \implies dx = \dfrac{du}{g'(x)}$.
  • Express the initial integral in terms of $u$ and $du$.
  • Evaluate the new integral with respect to $u$.
  • Replace $u$ with $g(x)$ in the final result.

EXAMPLE 1

Evaluate $I = \int 3x^2(x^3+5)^7 dx$.

Solution:

We let $u = x^3+5$ because the derivative $3x^2$ exists as a factor inside the integral.

  • Let $u = x^3+5$.
  • Then $\dfrac{du}{dx} = 3x^2 \implies dx = \dfrac{du}{3x^2}$.
  • Substitute these into the integral:
$$ I = \int 3x^2 u^7 \dfrac{du}{3x^2} = \int u^7 du $$

Evaluate the new integral:

$$ I = \dfrac{u^8}{8} + c $$

Replace $u$ with $x^3+5$:

$$ \mathbf{I = \dfrac{(x^3+5)^8}{8} + c} $$

EXAMPLE 2

Find $I = \int x\sqrt{x^2+3}dx$ by using the substitution $u=x^2+3$.

Solution:

Let $u=x^2+3$, then $\dfrac{du}{dx} = 2x \implies dx = \dfrac{du}{2x}$.

$$ \begin{aligned} I &= \int x \sqrt{u} \dfrac{du}{2x} = \dfrac{1}{2}\int u^{1/2} du = \dfrac{1}{2} \cdot \dfrac{u^{3/2}}{3/2} + c \\ &= \dfrac{1}{3}u^{3/2} + c = \mathbf{\dfrac{1}{3}(x^2+3)^{3/2} + c} \end{aligned} $$

EXAMPLE 3

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

Solution:

The derivative of the denominator is the numerator. Let $u = x^2+\sin x$, then $\dfrac{du}{dx} = 2x+\cos x \implies dx = \dfrac{du}{2x+\cos x}$.

$$ \begin{aligned} I &= \int \dfrac{2x+\cos x}{u} \dfrac{du}{2x+\cos x} \\ &= \int \dfrac{1}{u}du \\ &= \ln|u| + c \\ &= \mathbf{\ln|x^2+\sin x| + c} \end{aligned} $$

EXAMPLE 4

Find $I_1 = \int (\ln x)^2 \dfrac{1}{x} dx$, $I_2 = \int \dfrac{1}{x(\ln x)^2} dx$, and $I_3 = \int \dfrac{1}{x\ln x} dx$.

Solution:

For all three integrals, let $u = \ln x$. Then $\dfrac{du}{dx} = \dfrac{1}{x} \implies dx = x du$.

$$ I_1 = \int u^2 \dfrac{1}{x} x du = \int u^2 du = \dfrac{u^3}{3} + c = \mathbf{\dfrac{(\ln x)^3}{3} + c} $$
$$ I_2 = \int \dfrac{1}{x u^2} x du = \int u^{-2} du = \dfrac{u^{-1}}{-1} + c = -\dfrac{1}{u} + c = \mathbf{-\dfrac{1}{\ln x} + c} $$
$$ I_3 = \int \dfrac{1}{x u} x du = \int \dfrac{1}{u} du = \ln|u| + c = \mathbf{\ln|\ln x| + c} $$

EXAMPLE 5

A modification is needed to achieve the required form. Find $I = \int \dfrac{x}{x^2+4} dx$.

Solution:

The derivative of $x^2+4$ is $2x$, not $x$. We can adjust the integral:

$$ I = \dfrac{1}{2}\int \dfrac{2x}{x^2+4} dx $$

Let $u = x^2+4 \implies \dfrac{du}{dx} = 2x \implies dx = \dfrac{du}{2x}$.

$$ \begin{aligned} I &= \dfrac{1}{2} \int \dfrac{2x}{u} \dfrac{du}{2x} = \dfrac{1}{2} \int \dfrac{1}{u} du \\ &= \dfrac{1}{2}\ln|u| + c = \mathbf{\dfrac{1}{2}\ln(x^2+4) + c} \end{aligned} $$

EXAMPLE 6

Apply the substitution rule to evaluate the following integrals:

  • $\int e^{3x}dx = \mathbf{\dfrac{e^{3x}}{3} + c}$
  • $\int e^{-5x+2}dx = \mathbf{\dfrac{e^{-5x+2}}{-5} + c}$
  • $\int \sin(2x+1)dx = \mathbf{-\dfrac{\cos(2x+1)}{2} + c}$
  • $\int \cos(5-x)dx = \dfrac{\sin(5-x)}{-1} + c = \mathbf{-\sin(5-x) + c}$
  • $\int \dfrac{1}{7x+3}dx = \mathbf{\dfrac{\ln|7x+3|}{7} + c}$
  • $\int \dfrac{1}{x-3}dx = \mathbf{\ln|x-3| + c}$
  • $\int (3x+5)^5 dx = \dfrac{(3x+5)^6}{6 \cdot 3} + c = \mathbf{\dfrac{(3x+5)^6}{18} + c}$
  • $\int \dfrac{1}{(3x+5)^4} dx = \int (3x+5)^{-4} dx = \dfrac{(3x+5)^{-3}}{-3 \cdot 3} + c = \mathbf{-\dfrac{1}{9(3x+5)^3} + c}$
  • $\int \sqrt{3x+5} dx = \int (3x+5)^{1/2} dx = \dfrac{(3x+5)^{3/2}}{(3/2) \cdot 3} + c = \mathbf{\dfrac{2(3x+5)^{3/2}}{9} + c}$

2. Useful Trigonometric Formulas

$\sin^2 x = \dfrac{1-\cos(2x)}{2}$
$\cos^2 x = \dfrac{1+\cos(2x)}{2}$
$\sin(2x) = 2\sin x \cos x$
$\cos(2x) = \cos^2 x - \sin^2 x = 1 - 2\sin^2 x = 2\cos^2 x - 1$
$\sin^2 x + \cos^2 x = 1$
$1 + \tan^2 x = \dfrac{1}{\cos^2 x} = \sec^2 x$

EXAMPLE 7

Find $I = \int \cos^2 x dx$ by using the double angle formula $\cos^2 x = \dfrac{1+\cos(2x)}{2}$.

Solution:
$$ \begin{aligned} I &= \int \cos^2 x dx = \int \dfrac{1+\cos(2x)}{2} dx = \int \left( \dfrac{1}{2} + \dfrac{\cos(2x)}{2} \right) dx \\ &= \dfrac{1}{2}x + \dfrac{\sin(2x)}{2 \cdot 2} + c = \mathbf{\dfrac{x}{2} + \dfrac{\sin(2x)}{4} + c} \end{aligned} $$