# 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.

## Examples

### Example 1

Evaluate the integral
Solution to Example 1:
Let u = sin(x) and dv/dx = ex and then use the integration by parts as follows

We apply the integration by parts to the term cos(x)ex dx in the expression above, hence

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)e
x +2sin(2x)ex dx
Apply integration by parts to the term on the right
= cos(2x)e
x + 2{sin(2x)ex - 2cos(2x)ex dx }
= cos(2x)e
x + 2sin(2x)ex - 4cos(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)e
3x - {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) e
2x + 5 + 2{ sin(4x) (1/2)e2x + 5 - 2cos(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

1. (1/2) e
x {cos(x) + sin(x)} + C
2. (1/13)e
3x {3sin(2x) - 2cos(2x)} + C
3. (1/34) e
5x { 5cos(-3x + 5) - 3sin(-3x + 5) } + C
4. (1/10) e
-2x + 1 { 2cos(-4x + 3) - sin(-4x + 3) } + C