While integration by substitution, commonly referred to as u-substitution is a common and vital method for solving integrals in calculus. It makes the integration easier because it can convert a complicated integration into a more manageable one. This method is important for students as it builds up their knowledge to progress to other integration methods.
What is Integration by Substitution?
Integration by substitution is one of the techniques of definite integration by which we substitute the variable of integration with one or more new variables. This is a method applied to an integral whereby a certain section of the integrand is replaced by a new variable for instance u.
This technique becomes very helpful when the integrand to be integrated is a composite function and involves one function inside another. By making an appropriate substitution, the integral can be transformed, making it easier to solve.
Important/Related Formulas and Concepts
- Basic Substitution Formula: If u=g(x), then du=g′(x)dx. The integral
\int f(g(x)) g'(x) \, dx becomes\int f(u) \, du - Reversing Substitution: After integrating in terms of u, substitute u=g(x) to express the final answer in terms of x.
Steps for Integration by Substitution
Various steps for integration by substitution are:
Step 1: Identify the part of the integrand that can be substituted (usually a composite function).
Step 2: Define the substitution u=g(x).
Step 3: Compute du=g′(x)dx.
Step 4: Rewrite the integral in terms of u and du.
Step 5: Integrate with respect to u.
Step 6: Substitute back u=g(x) to get the result in terms of x.
Key Formulas:
Integration by Substitution Practice Problems
Problem 1: Evaluate the integral
Solution:
This integral can be solved directly.
\int (2x+3) \, dx = \int 2x \, dx + \int 3 \, dx
= 2 \int x \, dx + 3 \int 1 \, dx
= 2 \left(\frac{x^2}{2}\right) + 3x + C
= x^2 + 3x + C
Problem 2: Evaluate the integral
Solution:
Let u = x2
Then, du = 2xdx, and hence xdx = 1/2du
\int x \cos(x^2) \, dx = \int \cos(u) \cdot \frac{1}{2} \, du
= \frac{1}{2} \int \cos(u) \, du
= \frac{1}{2} \sin(u) + C
= \frac{1}{2} \sin(x^2) + C
Problem 3: Evaluate the integral
Solution:
Let u = 3x
Then, du = 3dx, and hence dx = 1/3 du
\int e^{3x} \, dx = \int e^u \cdot \frac{1}{3} \, du
= \frac{1}{3} \int e^u \, du
= \frac{1}{3} e^u + C
= \frac{1}{3} e^{3x} + C
Problem 4: Evaluate the integral
Solution:
Let u = ln(x)
Then, du = 1/x dx, and hence dx = xdu
\int \frac{1}{x \ln(x)} \, dx = \int \frac{1}{u} \cdot \frac{1}{x} x \, du
= \int \frac{1}{u} \, du
= \ln|u| + C
= \ln|\ln(x)| + C
Problem 5: Evaluate the integral
Solution:
Let u = x3
Then, du = 3x2 dx, and hence x2dx = 1/3 du
\int x^2 e^{x^3} \, dx = \int e^u \cdot \frac{1}{3} \, du
= \frac{1}{3} \int e^u \, du
= \frac{1}{3} e^u + C
= \frac{1}{3} e^{x^3} + C
Problem 6: Evaluate the integral
Solution:
Let u = 3x
Then, du = 3dx, and hence dx = 1/3 du
\int \cos(3x) \, dx = \int \cos(u) \cdot \frac{1}{3} \, du
= \frac{1}{3} \int \cos(u) \, du
= \frac{1}{3} \sin(u) + C
= \frac{1}{3} \sin(3x) + C
Problem 7: Evaluate the integral
Solution:
Let u = x2 + 1
Then, du = 2xdx, and hence xdx = 1/2 du
\int x \sqrt{x^2 + 1} \, dx = \int \sqrt{u} \cdot \frac{1}{2} \, du
= \frac{1}{2} \int u^{1/2} \, du
= \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C
= \frac{1}{3} (x^2 + 1)^{3/2} + C
Problem 8: Evaluate the integral
Solution:
Let u = ln(x)
Then, du = 1/x dx, and hence dx = xdu
\int \sin(\ln(x)) \, dx = \int \sin(u) x \, du Since x = eu,
\int \sin(\ln(x)) \, dx = \int \sin(u) e^u \, du We can solve this by integrating parts. Let v = eu, dv = eu du, and w = sin(u), dw = cos(u)du.
\int \sin(u) e^u \, du = -e^u \cos(u) + \int e^u \cos(u) \, du We solve the integral
\int e^u cos(u)du by parts again. Let v = eu, dv = eu du, and w = cos(u), dw = −sin(u)du.
\int e^u \cos(u) \, du = e^u \sin(u) + \int e^u \sin(u) \, du Thus,
\int e^u \sin(u) \, du = -e^u \cos(u) + e^u \sin(u) + C
= e^u (\sin(u) - \cos(u)) + C
= x (\sin(\ln(x)) - \cos(\ln(x))) + C
Problem 9: Evaluate the integral
Solution:
This is a standard integral that results in the arctangent function.
\int \frac{1}{1+x^2} \, dx = \arctan(x) + C
Problem 10: Evaluate the integral
Solution:
Let u = 1 − x2. Then, du = −2xdx, and hence xdx = − 1/2 du.
\int \frac{x}{\sqrt{1 - x^2}} \, dx = -\frac{1}{2} \int \frac{1}{\sqrt{u}} \, du
= -\frac{1}{2} \int \frac{1}{\sqrt{u}} \, du
= -\frac{1}{2} \cdot 2\sqrt{u} + C
= -\sqrt{1 - x^2} + C
Practice Problems on Integration by Substitution Practice Problems
P1. Evaluate the integral
P2. Find the value of
P3. Solve the integral
P4. Evaluate
P5. Determine the integral
P6. Calculate the integral
P7. Solve for
P8. Evaluate
P9. Find the integral
P10. Calculate
Answer Key
1.
\int x e^{x^2} \, dx = \frac{1}{2} e^{x^2} + C 2.
\int \frac{\ln(x)}{x} \, dx = \frac{(\ln(x))^2}{2} + C 3.
\int \sin(x) \cos(x) \, dx = \frac{1}{2} \sin^2(x) + C 4.
\int \frac{2x}{(1 + x^2)^2} \, dx = -\frac{1}{1 + x^2} + C 5.
\int x^3 \sqrt{x^2 + 5} \, dx = \frac{1}{5} (x^2 + 5)^{3/2} - \frac{1}{3} (x^2 + 5)^{3/2} + C 6.
\int e^{2x} \sin(3x) \, dx = \frac{e^{2x}}{13} \left(3 \sin(3x) - 2 \cos(3x)\right) + C 7.
\int \frac{x}{\sqrt{4 - x^2}} \, dx = -\sqrt{4 - x^2} + C 8.
\int \tan(x) \, dx = -\ln|\cos(x)| + C 9.
\int \frac{x^2}{(x^3 + 1)^2} \, dx = -\frac{1}{3(x^3 + 1)} + C 10.
\int \frac{1}{\sqrt{x^2 + 4}} \, dx = \sinh^{-1}\left(\frac{x}{2}\right) + C
Conclusion
Integration by substitution is one of the most popular methods in integration techniques used in calculus which enable the calculation of the integral by substituting the integrand with a simpler form. Therefore, by experiencing this method in solving problems, the student will be in a position to identify the tool required to solve more problems that involve integration in Integral Calculus and other topics in Mathematics. The problems indicated and discussed in this article have been selected to let the students revise the discussed technique more effectively and prepare for the possible integration tasks.