# 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 1Evaluate the integralSolution to Example 1:Let u = sin(x) and dv/dx = e ^{x} and then use the integration by parts as followsWe apply the integration by parts to the term ∫ cos(x)e ^{x} dx in the expression above, henceSimplify 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 2Evaluate the integral^{x} dx
Solution to Example 2:Substitution: Let u = cos(2x) and dv/dx = e ^{x} and apply the integration by parts.∫cos(2x)e ^{x} dx = cos(2x)e^{x} -∫-2sin(2x)e^{x} dx
= cos(2x)e ^{x} +∫2sin(2x)e^{x} dx
Apply integration by parts to the term on the right = cos(2x)e ^{x} + 2{sin(2x)e^{x} - 2∫cos(2x)e^{x} dx }
= cos(2x)e ^{x} + 2sin(2x)e^{x} - 4∫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∫cos(2x)e ^{x} dx = cos(2x)e^{x} + 2sin(2x)e^{x}The given integral is ∫cos(2x)e ^{x} dx = (1/5)e^{x} {cos(2x) + 2sin(2x)} + C
## Example 3Evaluate the integral^{3x} dx
Solution to Example 3:Substitution: Let u = sin(3x + 2) and dv/dx = e ^{3x} and apply the integration by parts twice.∫sin(3x + 2) e ^{3x} dx = sin(3x + 2) (1/3)e^{3x} -∫cos(3x + 2)e^{3x} dx
= (1/3) sin(3x + 2)e ^{3x} - {cos(3x + 2)(1/3)e^{3x} +
∫sin(3x + 2) e^{3x}dx}
Note that the term on the right is the integral to be evaluated, hence ∫sin(3x + 2) e ^{3x} dx = (1/6) e^{3x} { sin(3x + 2) - cos(3x + 2) } + C
## Example 4Evaluate the integral^{2x + 5} dx
Solution to Example 4:Substitution: Let u = cos(4x) and dv/dx = e ^{2x + 5} and apply the integration by parts twice.∫cos(4x) e ^{2x + 5} dx = cos(4x) (1/2) e^{2x + 5} + 2∫ sin(4x) e^{2x + 5} dx
= cos(4x) (1/2) e ^{2x + 5} + 2{ sin(4x) (1/2)e^{2x + 5} - 2∫cos(4x)e^{2x + 5} dx }
The term on the right is the integral to be evaluated, hence ∫cos(4x) e ^{2x + 5} dx = (1/10)e^{2x + 5} {cos(4x) + 2 sin(4x)} + C
## ExercisesEvaluate the following integrals.1. ∫cos(x) e ^{x} dx
2. ∫sin(2x) e ^{3x} x dx
3. ∫cos(-3x + 5) e ^{5x}) dx
4. ∫sin(-4x + 3) e ^{-2x + 1} dx
## Answers to Above Exercises1. (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
## More References and Linksintegrals and their applications in calculus. |