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 follows
We apply the integration by parts to the term ∫ cos(x)e ^{x} 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 2Evaluate the integral^{x} dxSolution 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} dxSolution 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} dxSolution 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. |