A tutorial, with examples and detailed solutions, in using the rules of indefinite integrals in calculus is presented. A set of questions with solutions is also included.
In what follows, \( C \) is a constant of integration and can take any value.
1 - Integral of a power function: \( f(x) = x^n \)
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + c \]Example: Evaluate the integral \( \displaystyle \int x^5 \, dx \)
Solution:
\[ \int x^5 \, dx = \dfrac{x^{5 + 1}}{5 + 1} + c = \dfrac{x^6}{6} + c \]2 - Integral of a function \( f \) multiplied by a constant \( k \): \( k f(x) \)
\[ \int k f(x) \, dx = k \int f(x) \, dx \]Example: Evaluate the integral \( \displaystyle \int 5 \sin(x) \, dx \)
Solution:
According to the above rule:
\[ \int 5 \sin(x) \, dx = 5 \int \sin(x) \, dx \]\( \displaystyle \int \sin(x) \, dx \) is given by formula 2.1 in the table of integral formulas, hence:
\[ \int 5 \sin(x) \, dx = -5 \cos x + C \]3 - Integral of Sum of Functions
\[ \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \]Example: Evaluate the integral \( \displaystyle \int (x + e^x) \, dx \)
Solution:
According to the above property:
\[ \int (x + e^x) \, dx = \int x \, dx + \int e^x \, dx \]\( \int x \, dx \) is given by 1.3 and \( \displaystyle \int e^x \, dx \) by 4.1 in the table of integral formulas, hence:
\[ \int (x + e^x) \, dx = \dfrac{x^2}{2} + e^x + c \]4 - Integral of Difference of Functions
\[ \int (f(x) - g(x)) \, dx = \int f(x) \, dx - \int g(x) \, dx \]Example: Evaluate the integral \( \displaystyle \int \left(2 - \frac{1}{x}\right) \, dx \)
Solution:
According to the above property:
\[ \int \left(2 - \frac{1}{x}\right) dx = \int 2 \, dx - \int \frac{1}{x} \, dx \]\( \int 2 \, dx \) is given by 1.2 and \( \int \frac{1}{x} \, dx \) by 1.4 in the table of integral formulas, hence:
\[ \int \left(2 - \frac{1}{x}\right) dx = 2x - \ln|x| + c \]5 - Integration by Substitution
\[ \int f(u) \dfrac{du}{dx} \, dx = \int f(u) \, du \]Example: Evaluate the integral \( \displaystyle \int (x^2 - 1)^{20} \cdot 2x \, dx \)
Solution:
Let \( u = x^2 - 1 \), hence \( \dfrac{du}{dx} = 2x \) and the given integral can be written as:
\[ \int (x^2 - 1)^{20} \cdot 2x \, dx = \int u^{20} \dfrac{du}{dx} \, dx = \int u^{20} \, du \]Which evaluates to:
\[ = \dfrac{u^{21}}{21} + c \]Substitute back \( u = x^2 - 1 \):
\[ = \dfrac{(x^2 - 1)^{21}}{21} + c \]6 - Integration by Parts
\[ \int f(x) g'(x) \, dx = f(x)g(x) - \int f'(x) g(x) \, dx \]Example: Evaluate the integral \( \displaystyle \int x \cos x \, dx \)
Solution:
Let \( f(x) = x \) and \( g'(x) = \cos x \), which gives \( f'(x) = 1 \) and \( g(x) = \sin x \).
From the integration by parts formula above:
\[ \int x \cos x \, dx = x \sin x - \int 1 \cdot \sin x \, dx \] \[ = x \sin x + \cos x + c \]More Questions with Solutions
Use the table of integral formulas and the rules above to evaluate the following integrals. [Note that you may need to use more than one of the above rules for one integral].
Click on each question to view its detailed step-by-step solution.
Question 1: Evaluate \( \displaystyle \int \frac{1}{2} \ln(x) \, dx \)
Solution:
Take the constant \( \frac{1}{2} \) outside the integral:
\[ \int \frac{1}{2} \ln(x) \, dx = \frac{1}{2} \int \ln(x) \, dx \]Use integration by parts. Let \( u = \ln(x) \) and \( dv = dx \), then \( du = \frac{1}{x} \, dx \) and \( v = x \):
\[ \int u \, dv = u v - \int v \, du \] \[ \int \ln(x) \, dx = x\ln(x) - \int x \cdot \frac{1}{x} \, dx = x\ln(x) - \int 1 \, dx = x\ln(x) - x + C \]Therefore:
\[ \int \frac{1}{2} \ln(x) \, dx = \frac{1}{2}\bigl(x\ln(x) - x\big) + C \]Question 2: Evaluate \( \displaystyle \int (\sin x + x^5) \, dx \)
Solution:
Use rule 3 (integral of a sum) to obtain:
\[ \int (\sin x + x^5) \, dx = \int \sin x \, dx + \int x^5 \, dx \]We use formula 2.1 in the table of integral formulas to evaluate \( \int \sin x \, dx \) and rule 1 above to evaluate \( \int x^5 \, dx \). Hence:
\[ \int (\sin x + x^5) \, dx = -\cos x + \frac{x^6}{6} + c \]Question 3: Evaluate \( \displaystyle \int (\sinh x - 3) \, dx \)
Solution:
Use rule 4 (integral of a difference) to obtain:
\[ \int (\sinh x - 3) \, dx = \int \sinh x \, dx - \int 3 \, dx \]We use formula 7.1 in the table of integral formulas to evaluate \( \int \sinh x \, dx \) and the integral of the constant \( 3 \) to obtain:
\[ \int (\sinh x - 3) \, dx = \cosh x - 3x + c \]Question 4: Evaluate \( \displaystyle \int x \sin x \, dx \)
Solution:
The integrand is the product of two functions \( x \) and \( \sin x \). We use integration by parts (rule 6):
Let \( f(x) = x \) and \( g'(x) = \sin x \), hence \( f'(x) = 1 \) and \( g(x) = -\cos x \).
Then:
\[ \int x \sin x \, dx = f(x)g(x) - \int f'(x)g(x) \, dx = -x \cos x - \int 1 \cdot (-\cos x) \, dx = -x \cos x + \int \cos x \, dx \]Using formula 2.2 in the table of integral formulas to evaluate \( \int \cos x \, dx \), we obtain:
\[ \int x \sin x \, dx = -x \cos x + \sin x + c \]Question 5: Evaluate \( \displaystyle \int \sin^{10}(x) \cos(x) \, dx \)
Solution:
Let \( u = \sin x \), therefore \( du = \cos x \, dx \). Hence the given integral can be written as:
\[ \int \sin^{10} x \cos x \, dx = \int u^{10} \, du \]Use rule 1 to write:
\[ \int u^{10} \, du = \frac{u^{11}}{11} + c \]Substitute \( u = \sin x \) to obtain:
\[ \int \sin^{10} x \cos x \, dx = \frac{1}{11} \sin^{11} x + c \]