Integration Techniques: (lesson 1 of 4)
Integration by Substitution
The substitution method turns an unfamiliar integral into one that can be evaluated.
In other words, substitution gives a simpler integral involving the variable u u u
Let's review the five steps for integration by substitution.
Step 1: Choose a new variable u u u
Step 2: Determine the value d x dx d x
Step 3: Make the substitution
Step 4: Integrate resulting integral
Step 5: Return to the initial variable x
Integration by substituting u = a x + b u = ax + b u = a x + b
These are typical examples where the method of substitution is used.
Example 1: Solve:
∫ ( 2 x + 3 ) 4 d x
\int {(2x + 3)^4dx}
∫ ( 2 x + 3 ) 4 d x Solution:
Step 1: Choose the substitution function u u u
The substitution function is u = 2 x + 3 \color{blue}{u = 2x + 3} u = 2 x + 3
Step 2: Determine the value of d x dx d x
2 x + 3 = u ( 2 x + 3 ) ′ d x = d u 2 d x = d u d x = 1 2 d u
\begin{aligned}
2x + 3 &= u \\
(2x + 3)'dx &= du \\
2dx &= du \\
\color{red}{dx} &\color{red}{= \frac{1}{2}du}
\end{aligned}
2 x + 3 ( 2 x + 3 ) ′ d x 2 d x d x = u = d u = d u = 2 1 d u
Step 3: Do the substitution
∫ ( 2 x + 3 ) 4 d x = ∫ u 4 ⋅ 1 2 ⋅ d u
\int {\color{blue}{(2x + 3)}^4 dx} = \int {\color{blue}{u}^4} \cdot \color{red}{\frac{1}{2} \cdot du}
∫ ( 2 x + 3 ) 4 d x = ∫ u 4 ⋅ 2 1 ⋅ d u
Step 4: Integrate the resulting integral
∫ ( 2 x + 3 ) 4 d x = ∫ u 4 ⋅ 1 2 ⋅ d u = 1 2 ∫ u 4 d u = 1 2 ⋅ u 4 + 1 4 + 1 + C = u 5 10 + C
\int {{{(2x + 3)}^4}} dx = \int {{u^4}} \cdot \frac{1}{2} \cdot du \color{red}{=} \frac{1}{2}\int {{u^4}} du = \frac{1}{2} \cdot \frac{{{u^{4 + 1}}}}{{4 + 1}} + C = \frac{{{u^5}}}{{10}} + C
∫ ( 2 x + 3 ) 4 d x = ∫ u 4 ⋅ 2 1 ⋅ d u = 2 1 ∫ u 4 d u = 2 1 ⋅ 4 + 1 u 4 + 1 + C = 10 u 5 + C
Step 5: Return to the initial variable: x x x
u 5 10 + C = ( 2 x + 3 ) 5 10 + C
\frac{\color{blue}{u}^5}{10} + C = \frac{\color{blue}{(2x + 3)}^5}{10} + C
10 u 5 + C = 10 ( 2 x + 3 ) 5 + C
So, the solution is:
∫ ( 2 x + 3 ) 4 d x = ( 2 x + 3 ) 5 10 + C \int (2x + 3) ^ 4 dx = \frac{(2x + 3)^5}{10} + C ∫ ( 2 x + 3 ) 4 d x = 10 ( 2 x + 3 ) 5 + C
Example 2: Solve:
∫ 15 3 − 2 x d x \int {\frac{15}{3 - 2x}dx} ∫ 3 − 2 x 15 d x Solution:
Step 1: Choose a substitution function u u u
The substitution function is
u = 3 − 2 x \color{blue}{u = 3 - 2x} u = 3 − 2 x
Step 2: Determine the value d x dx d x
3 − 2 x = u ( 3 − 2 x ) ′ d x = d u − 2 d x = d u d x = − 1 2 d u
\begin{aligned}
3 - 2x &= u \\
(3 - 2x)' dx &= du \\
-2dx &= du \\
\color{red}{dx} &\color{red}{= - \frac{1}{2} du}
\end{aligned}
3 − 2 x ( 3 − 2 x ) ′ d x − 2 d x d x = u = d u = d u = − 2 1 d u
Step 3: Do the substitution
∫ 15 3 − 2 x d x = ∫ 15 u ⋅ ( − 1 2 ) ⋅ d u
\int {\frac{15}{\color{blue}{3 - 2x}} \color{red}{dx}} = \int {\frac{15}{\color{blue}{u}} \cdot \color{red}{\left( { - \frac{1}{2}} \right) \cdot du}}
∫ 3 − 2 x 15 d x = ∫ u 15 ⋅ ( − 2 1 ) ⋅ d u
Step 4: Integrate resulting integral
∫ 15 u ⋅ ( − 1 2 ) ⋅ d u = − 15 2 ∫ 1 u d u = − 15 2 ln ∣ u ∣ + C
\int {\frac{15}{u} \cdot \left( { - \frac{1}{2}} \right) \cdot du = - \frac{15}{2}\int {\frac{1}{u}du = - \frac{15}{2}\ln \left| u \right|} + C}
∫ u 15 ⋅ ( − 2 1 ) ⋅ d u = − 2 15 ∫ u 1 d u = − 2 15 ln ∣ u ∣ + C
Step 5: Return to the initial variable: x x x
− 15 2 ln ∣ u ∣ + C = − 15 2 ln ∣ 3 − 2 x ∣ + C
- \frac{{15}}{2}\ln \left| \color{blue}{u} \right| + C = - \frac{{15}}{2}\ln \left| \color{blue}{3 - 2x} \right| + C
− 2 15 ln ∣ u ∣ + C = − 2 15 ln ∣ 3 − 2 x ∣ + C The solution is: ∫ 15 3 − 2 x d x = − 15 2 ln ∣ 3 − 2 x ∣ + C \int {\frac{{15}}{{3 - 2x}}dx = - \frac{{15}}{2}\ln \left| {3 - 2x} \right|} + C ∫ 3 − 2 x 15 d x = − 2 15 ln ∣ 3 − 2 x ∣ + C
Exercise 1: Solve using substitution u = a x + b u = ax + b u = a x + b
More complicated examples
The steps for integration by substitution in this section are the same as
the steps for previous one, but make sure to choose the substitution function wisely.
Example 3: Solve:
∫ x sin ( x 2 ) d x
\int {x\sin ({x^2})dx}
∫ x sin ( x 2 ) d x Solution:
Let u = x 2 then: d u = ( x 2 ) ′ d x = 2 x d x ⇒ x d x = d u 2 ∫ x sin ( x 2 ) d x = ∫ sin u ⋅ d u 2 = 1 2 ∫ sin u d u = − 1 2 cos u + C = − 1 2 cos x 2 + C
\text{Let} \ \ \color{blue}{u = x^2} \ \ \text{then:} \ \ du = (x^2)' dx = 2xdx \ \Rightarrow \ \color{red}{xdx = \frac{du}{2}} \\
\int {\color{red}{x} \sin (\color{blue}{x^2}) \color{red}{dx} = } \int {\sin \color{blue}{u} \cdot \color{red}{\frac{{du}}{2}}} = \frac{1}{2}\int {\sin udu = - \frac{1}{2}\cos u + C = - \frac{1}{2}\cos {x^2} + C}
Let u = x 2 then: d u = ( x 2 ) ′ d x = 2 x d x ⇒ x d x = 2 d u ∫ x s i n ( x 2 ) d x = ∫ s i n u ⋅ 2 d u = 2 1 ∫ s i n u d u = − 2 1 c o s u + C = − 2 1 c o s x 2 + C
Example 4: Solve:
∫ x 2 1 + x 3 d x
\int {{x^2}} \sqrt {1 + {x^3}} dx
∫ x 2 1 + x 3 d x Solution:
Let u = 1 + x 3 then: d u = ( 1 + x 3 ) ′ d x = 3 x 2 d x ⇒ x d x = d u 3 ∫ x 2 1 + x 3 d x = ∫ u ⋅ d u 3 = 1 3 ∫ u d u = = 1 3 ∫ u 1 2 d u = 1 3 ⋅ u 1 2 + 1 1 2 + 1 + C = 1 3 ⋅ 2 3 ⋅ u 3 2 + C = 2 9 ⋅ ( 1 + x 3 ) 3 2 + C
\text{Let} \ \ \color{blue}{u = 1 + x^3} \ \ \text{then:} \ \ du = (1 + x^3)' dx = 3x^2dx \ \Rightarrow \ \color{red}{x^dx = \frac{du}{3}} \\
\int {\color{red}{x^2}} \sqrt {\color{blue}{1 + x^3}} \color{red}{dx}
= \int {\color{blue}{\sqrt u} \cdot \color{red}{\frac{du}{3}} = \frac{1}{3}\int {\sqrt u du} = \\
= \frac{1}{3}\int {{u^{\frac{1}{2}}}} } du = \frac{1}{3} \cdot \frac{{{u^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}} + C = \frac{1}{3} \cdot \frac{2}{3} \cdot {\color{blue}{u}^{\frac{3}{2}}} + C = \frac{2}{9} \cdot {\color{blue}{(1 + x^3)}^{\frac{3}{2}}} + C
Let u = 1 + x 3 then: d u = ( 1 + x 3 ) ′ d x = 3 x 2 d x ⇒ x d x = 3 d u ∫ x 2 1 + x 3 d x = ∫ u ⋅ 3 d u = 3 1 ∫ u d u = = 3 1 ∫ u 2 1 d u = 3 1 ⋅ 2 1 + 1 u 2 1 + 1 + C = 3 1 ⋅ 3 2 ⋅ u 2 3 + C = 9 2 ⋅ ( 1 + x 3 ) 2 3 + C
Example 5: Solve:
∫ 1 x 2 ( 1 + 1 x ) 2 d x
\int {\frac{1}{{{x^2}{{\left( {1 + \frac{1}{x}} \right)}^2}}}} dx
∫ x 2 ( 1 + x 1 ) 2 1 d x Solution:
Let u = 1 + 1 x then: d u = ( 1 + 1 x ) ′ d x = − x − 2 = − 1 x 2 d x ∫ 1 x 2 ( 1 + 1 x ) d x = − ∫ 1 − x 2 ( 1 + 1 x ) 2 d x = − ∫ 1 u 2 d u = = − ∫ u − 2 d u = − u − 2 + 1 − 2 + 1 + C = u − 1 + C = ( 1 + 1 x ) − 1 + C = = ( x + 1 x ) − 1 + C = x x + 1 + C
\text{Let} \ \ \color{blue}{u = 1 + \frac{1}{x}} \ \ \text{then:} \ \color{red}{du} = \left( 1 + \frac{1}{x} \right)' dx = -x^{-2} = \color{red}{- \frac{1}{x^2} dx} \\
\int {\frac{1}{{{x^2}\left( {1 + \frac{1}{x}} \right)}}} dx = - \int {\frac{\color{red}{1}}{{ \color{red}{- {x^2}}{{\left( {\color{blue}{1 + \frac{1}{x}}} \right)}^2}}}} \color{red}{dx} = - \int {\frac{1}{\color{blue}{u^2}}} \color{red}{du} = \\
= - \int {{u^{ - 2}}du = - \frac{{{u^{ - 2 + 1}}}}{{ - 2 + 1}} + C = \color{blue}{u}^{ - 1} + C} = {\left( \color{blue}{1 + \frac{1}{x}} \right)^{ - 1}} + C = \\
= \left( \frac{x + 1}{x} \right)^{-1} + C = \frac{x}{x + 1} + C
Let u = 1 + x 1 then: d u = ( 1 + x 1 ) ′ d x = − x − 2 = − x 2 1 d x ∫ x 2 ( 1 + x 1 ) 1 d x = − ∫ − x 2 ( 1 + x 1 ) 2 1 d x = − ∫ u 2 1 d u = = − ∫ u − 2 d u = − − 2 + 1 u − 2 + 1 + C = u − 1 + C = ( 1 + x 1 ) − 1 + C = = ( x x + 1 ) − 1 + C = x + 1 x + C
Exercise 2: Solve using substitution
