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.

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 dv/dx = ex and then use the integration by parts as follows
\int \sin(x)e^x dx = \sin(x) e^x - \int \cos(x)e^x dx

We apply the integration by parts to the term ∫ cos(x)ex dx in the expression above, hence
\int sin(x) e^x dx = \sin(x) e^x - ( \cos(x)e^x - \int - \sin(x) e^x dx )

Simplify the above and rewrite as
\int sin(x) e^x dx = \sin(x) e^x - \cos(x)e^x - \int \sin(x) e^x dx

Note that the term on the right is the integral we are trying to evaluate, hence the above may be written as follows
2 \int sin(x) e^x dx = \sin(x) e^x - \cos(x)e^x
Hence the integral is given by
\int sin(x) e^x dx = \dfrac{1}{2} e^x ( \sin(x) - \cos(x))


Example 2: Evaluate the integral

∫cos(2x)ex dx

Solution to Example 2:
Substitution: Let u = cos(2x) and dv/dx = ex and apply the integration by parts.
∫cos(2x)ex dx = cos(2x)ex -∫-2sin(2x)ex dx
= cos(2x)ex +∫2sin(2x)ex dx
Apply integration by parts to the term on the right
= cos(2x)ex + 2{sin(2x)ex - 2∫cos(2x)ex dx }
= cos(2x)ex + 2sin(2x)ex - 4∫cos(2x)ex dx
Note that the term on the right is related to the integral we are trying to evaluate, we can write that
5∫cos(2x)ex dx = cos(2x)ex + 2sin(2x)ex
The given integral is
∫cos(2x)ex dx = (1/5)ex {cos(2x) + 2sin(2x)} + C

Example 3: Evaluate the integral

∫sin(3x + 2) e3x dx

Solution to Example 3:
Substitution: Let u = sin(3x + 2) and dv/dx = e3x and apply the integration by parts twice.
∫sin(3x + 2) e3x dx = sin(3x + 2) (1/3)e3x -∫cos(3x + 2)e3x dx
= (1/3) sin(3x + 2)e3x - {cos(3x + 2)(1/3)e3x + ∫sin(3x + 2) e3xdx}
Note that the term on the right is the integral to be evaluated, hence
∫sin(3x + 2) e3x dx = (1/6) e3x { sin(3x + 2) - cos(3x + 2) } + C

Example 4: Evaluate the integral

∫cos(4x) e2x + 5 dx

Solution to Example 4:
Substitution: Let u = cos(4x) and dv/dx = e2x + 5 and apply the integration by parts twice.
∫cos(4x) e2x + 5 dx = cos(4x) (1/2) e2x + 5 + 2∫ sin(4x) e2x + 5 dx
= cos(4x) (1/2) e2x + 5 + 2{ sin(4x) (1/2)e2x + 5 - 2∫cos(4x)e2x + 5 dx }
The term on the right is the integral to be evaluated, hence
∫cos(4x) e2x + 5 dx = (1/10)e2x + 5 {cos(4x) + 2 sin(4x)} + C

Exercises: Evaluate the following integrals.
1. ∫cos(x) ex dx
2. ∫sin(2x) e3x x dx
3. ∫cos(-3x + 5) e5x) dx
4. ∫sin(-4x + 3) e-2x + 1 dx

Answers to Above Exercises
1. (1/2) ex {cos(x) + sin(x)} + C
2. (1/13)e3x {3sin(2x) - 2cos(2x)} + C
3. (1/34) e5x { 5cos(-3x + 5) - 3sin(-3x + 5) } + C
4. (1/10) e-2x + 1 { 2cos(-4x + 3) - sin(-4x + 3) } + C
More references on integrals and their applications in calculus.