Solving Trig Derivatives with Integration by Parts

In summary, the integral of e^2x sin3x dx can be found by using the integration by parts formula twice to get 1/13 e^2x (2sin3x - 3cos3x).
  • #1
Wonderballs
31
0

Homework Statement



[tex]\int[/tex]e[tex]^{2x}[/tex]sin(3x)dx

Homework Equations



integration by parts formula


The Attempt at a Solution



= 9/10e[tex]^{2x}[/tex](sin(3x)-1/3cos(3x))

I don't think I am taking the derivative of sin(3x) correctly

thanks for taking a look :)
 
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  • #2
[tex]\int e^{2x}sin(3x)dx[/tex]

u=e^2x so that du=2e^2x dx

dv=sin(3x)dx ; v=-1/3 cos(3x)


but either way you would have to do integration by parts twice so I am thinking that is not the answer(BUT I can be wrong as I did not really work it out)
 
  • #3
Hint :

What you do is you integrate it by parts two times.

A multiple of the original integral will appear in the answer.
 
  • #4
I tried it with u = e^2x ; dv = sin3xdx

I think i got the multiple and I am stuck here:

= -1/3e[tex]^{2x}cos(3x) + 4/9[/tex][tex]\int[/tex]sin(3x)e[tex]^{2x}[/tex]
 
  • #5
Wonderballs said:
I tried it with u = e^2x ; dv = sin3xdx

I think i got the multiple and I am stuck here:

= -1/3e[tex]^{2x}cos(3x) + 4/9[/tex][tex]\int[/tex]sin(3x)e[tex]^{2x}[/tex]

you are supposed to get
[tex]\int e^{2x}sin(3x)dx=\frac{-e^{2x}}{3}cos(3x)+ \frac{2}{3}\int e^{2x}cos(3x)dx[/tex]

then do integration by parts again
 
  • #6
I got it, thanks guys

1/13 e^2x (2sin3x - 3cos3x) = integral
 

1. What is integration by parts?

Integration by parts is a method used in calculus to find the integral of a product of two functions. It is based on the product rule for differentiation and involves splitting the integral into two parts and using a specific formula to solve it.

2. How is integration by parts used to solve trigonometric derivatives?

To solve trigonometric derivatives using integration by parts, you need to identify which function will be integrated and which will be differentiated. You then use the formula: ∫u dv = uv - ∫v du to solve the integral, where u is the function being differentiated and dv is the function being integrated.

3. What are the steps to solve trig derivatives with integration by parts?

The steps to solve trig derivatives with integration by parts are:1. Identify which function will be differentiated and which will be integrated.2. Use the formula ∫u dv = uv - ∫v du to set up the integral.3. Differentiate the function u and integrate the function v.4. Substitute the values into the formula and solve for the integral.5. Simplify the answer and add any necessary constants.

4. What are some common mistakes to avoid when using integration by parts to solve trig derivatives?

Some common mistakes to avoid when using integration by parts to solve trig derivatives include:- Forgetting to use the formula ∫u dv = uv - ∫v du- Integrating the wrong function or differentiating the wrong function- Forgetting to apply the chain rule when differentiating- Forgetting to add the constant of integration when solving the integral- Making sign errors or algebraic mistakes during the calculation process

5. Are there any alternative methods for solving trig derivatives besides integration by parts?

Yes, there are alternative methods for solving trig derivatives, such as using the product rule, quotient rule, or chain rule. However, integration by parts is often a preferred method as it can simplify the calculation process and can be used for more complex integrals.

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