Integrals Involving sin x , cos x and Exponential Functions
Tutorial to find integrals involving the product of sin x or cos x with exponential functions. Exercises with answers are at the bottom of the page.
All the integrals included in the examples belo are evaluated using Integration by Parts given by:
The integration by parts helps in evaluating integral of product of functions of the form U dV/dx .
Examples
In what follows, C is the constant of integration.
Example 1
Evaluate the integral
\[ \int \sin(x) e^x \, dx \]
Solution to Example 1:
Let \( u = \sin(x) \) and \( \dfrac{dv}{dx} = e^x \) which gives \( u' = \cos(x) \) and \( v = \displaystyle \int e^x \, dx = e^x \).
Use the integration by parts as follows
We apply the integration by parts to the term \( \displaystyle \int \cos(x)e^x \, dx \) in the expression above, hence
Simplify the above and rewrite as
Note that the term on the right is the integral we are trying to evaluate, hence the above may be written as follows
Hence the integral is given by
Example 2
Evaluate the integral
\[ \int \cos(2x) \, e^x \, dx \]
Solution to Example 2:
Substitution: Let \( u = \cos(2x) \) and \( \dfrac{dv}{dx} = e^x \) which gives \( u' = - 2 \sin(2x) \) and \( v = \displaystyle \int e^x \, dx = e^x \) :
Apply the integration by parts:
\( \displaystyle \int \cos(2x) e^x \, dx = \cos(2x) \, e^x - \int -2 \sin(2x) e^x \, dx \)
= \( \cos(2x) \, e^x + 2 \int \sin(2x) \, e^x \, dx \)
Apply integration by parts to the term on the right
= \( \cos(2x) \, e^x + 2 ( \sin(2x)e^x - 2 \displaystyle \int \cos(2x)e^x \, dx ) \)
= \( \cos(2x) \, e^x + 2 \sin(2x) \, e^x - 4 \displaystyle \int \cos(2x)e^x \, dx \)
Note that the term on the right is related to the integral we are trying to evaluate, we can write that
\( 5 \displaystyle \int \cos(2x)e^x \, dx = \cos(2x) \, e^x + 2 \sin(2x) \, e^x \)
The given integral is evaluated as
\[ \int \cos(2x) \, e^x \, dx = \dfrac{1}{5} e^x ( \cos(2x) + 2 \sin(2x) ) + C \]
Example 3
Evaluate the integral
\[ \int \sin(3x + 2) \, e^{3x} \, dx \]
Solution to Example 3:
Substitution: Let \( u = \sin(3x + 2) \) and \( \dfrac{dv}{dx} = e^{3x} \) which gives \( u' = 3 \cos(3x + 2) \) and \( v = \displaystyle \int e^{3x} \, dx = \dfrac{1}{3} e^{3x} \).
Apply the integration by parts
\( \displaystyle \int \sin(3x + 2) \, e^{3x} \, dx = \sin(3x + 2) \dfrac{1}{3} e^{3x} - \int \cos(3x + 2) \dfrac{1}{3} e^{3x} \, dx \)
Apply the integration by parts one more time to the term \( \int \cos(3x + 2) e^{3x} \, dx \)
\( \displaystyle \int \sin(3x + 2) e^{3x} \, dx = \dfrac{1}{3} \sin(3x + 2) e^{3x} - ( \cos(3x + 2) \dfrac{1}{3} e^{3x} + \int \sin(3x + 2) e^{3x} \, dx ) \)
Note that the term on the right is the integral to be evaluated, hence the above may be written as
\( 2 \displaystyle \int \sin(3x + 2) e^{3x} \, dx = \dfrac{1}{3} \sin(3x + 2) e^{3x} = \dfrac{1}{3} \sin(3x + 2) e^{3x} - ( \cos(3x + 2)\dfrac{1}{3} e^{3x} \)
Divide all terms by 2 and simplify
\[ \int \sin(3x + 2) e^{3x} \, dx = \dfrac{1}{6} e^{3x} ( \sin(3x + 2) - \cos(3x + 2) ) + C \]
Example 4
Evaluate the integral
\[ \int \cos(4x) \, e^{2x + 5} \, dx \]
Solution to Example 4:
Substitution: Let \( u = \cos(4x) \) and \( \dfrac{dv}{dx} = e^{2x + 5} \) and apply the integration by parts twice
\( \displaystyle \int \cos(4x) \, e^{2x + 5} \, dx = \cos(4x) \dfrac{1}{2} e^{2x + 5} + 2 \int \sin(4x) \, e^{2x + 5} \, dx \)
= \( \displaystyle \cos(4x) \dfrac{1}{2} e^{2x + 5} + 2 \{ \sin(4x) \dfrac{1}{2}e^{2x + 5} - 2 \int \cos(4x) \, e^{2x + 5} \, dx \} \)
The term on the right is the integral to be evaluated, hence
\[ \int \cos(4x) \, e^{2x + 5} \, dx = \dfrac{1}{10} e^{2x + 5} ( \cos(4x) + 2 \sin(4x) ) + C \]
Exercises
Evaluate the following integrals.
1. \( \displaystyle\int \cos(x) \, e^x \, dx \)
2. \( \displaystyle \int \sin(2x) \, e^{3x} \, dx \)
3. \( \displaystyle \int \cos(-3x + 5) \, e^{5x} \, dx \)
4. \( \displaystyle \int \sin(-4x + 3) \, e^{-2x + 1} \, dx \)
Answers to Above Exercises
1. \( \dfrac{1}{2} e^x ( \cos(x) + \sin(x) ) + C \)
2. \( \dfrac{1}{13} e^{3x} ( 3 \sin(2x) - 2 \cos(2x)) + C \)
3. \( \dfrac{1}{34} e^{5x} ( 5 \cos(-3x + 5) - 3 \sin(-3x + 5) ) + C \)
4. \( \dfrac{1}{10} e^{-2x + 1} ( 2 \cos(-4x + 3) - \sin(-4x + 3) ) + C \)
More References and links