Need Help with Integration by Parts? Check Out These Tricky Problems!

In summary: It can also be expanded to LIATE (Logarithmic, Inverse trig., Algebraic, Trig., Exponential) to include more types of functions. But for this problem, using the product to sum rule is more efficient and appropriate. Now, for the summary:In summary, the conversation is about a person seeking help with integration by parts problems in calculus. Three specific problems are mentioned: finding the integral of (e^2x)sin3xdx, cos(3x+1)cos(5x+6)dx, and x²ln²xdx. Various methods and techniques are suggested, such as using u-substitution and the product to sum rule. The conversation ends with the person thanking the others for
  • #1
Integralien
4
0
Integration by parts HELP!

Hey, I am working on some calculus and I am having some trouble with the last few integration by parts problems. I got the first couple, and i grasp the concept of integration by parts but for some reason I just can't figure these 3. Any help would be greatly appreciated.

1. The integral of (e^2x)sin3xdx

2. The integral of cos(3x+1)cos(5x+6)dx

3. The integral of x²ln²xdx

P.S> Feel free to ask if you can't read the questions, I tried to write them out the best I could.


Thanks in advance,

Brian
 
Last edited:
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  • #2
1) u=e^2x ; dv=sin3x dx

du=2e^2x dx ; v=[itex]\frac{-1}{3}cos3x[/itex]

[tex]\int e^{2x}sin3x dx= \frac{-e^{2x}}{3}cos3x-\int \frac{-2e^{2x}}{3}cos3x dx[/tex]

repeat in the same fashion
 
  • #3
3) [tex]\int x^{2}ln^{2}xdx[/tex] now let
[tex]u=ln^{2}x \ so \ du=\frac{2lnx}{x} \ , \ v=\int x^{2}dx=\frac{x^{3}}{3}[/tex] now:
[tex]\frac{x^{3}}{3}ln^{2}x-\int \frac{2lnx \ x^{3}}{3x}dx=\frac{x^{3}}{3}ln^{2}x-\frac{2}{3}\int x^{2}lnxdx[/tex] now repeat integraion by part also for

[tex]\int x^{2}lnx dx[/tex], and i think you will get the answer!
 
  • #4
i am going to give u a hint on the 2) also
[tex]I=\int cos(3x+1)cos(5x+6)dx[/tex] now let[tex]u=cos(3x+1) \ => \ du=-3sin(3x+1) \ and \ \ v=\int cos(5x+6)dx=\frac{sin(5x+6}{5}[/tex] so now we have:

[tex]\frac{cos(3x+1)sin(5x+6)}{5}+\frac{3}{5}\int sin(3x+1)sin(5x+6)dx[/tex]

now for the integral [tex] \int sin(3x+1)sin(5x+6)dx[/tex] take this sub.

[tex]u=sin(3x+1) \ => du=3cos(3x+1) \ \, and \ \ v=\int sin(5x+6)dx = -\frac{cos(5x+6)}{5}[/tex] now u have:
[tex] -\frac{sin(3x+1)cos(5x+6)}{5} +\frac{3}{5} \int cos(3x+1)cos(5x+6)dx[/tex]
Now do you see anything interesting in here, anything that should grab your attention, try to go from here, because i think i have almost done it, you are almost there!@!
 
  • #5
Thanks a lot guys I believe I have come to the right answers. Thanks a million!
 
  • #6
For number 2.

How about using the product to sum rule cos(A)cos(B)=1/2[cos(A-B)+cos(A+B)]

then cos(3x+1)cos(5x+6)=1/2[cos(8x+7)+cos(2x+5)]

then integrate from here.
 
  • #7
Not sure if your calculus teacher ever taught you this neat mnemonic, so I'll share it anyways: LIPET, which stands for Logarithmic, Inverse trig., Polynomial, Exponential, and Trigonometric. This is the order in which you should choose for something to set as "u" when integrating by parts. If you've already heard this, sorry I can't be of more help, just thought I'd add on something since you were talking about integration by parts.
 
  • #8
LIPET is very useful and worth memorizing.
 

Related to Need Help with Integration by Parts? Check Out These Tricky Problems!

1. What is integration by parts?

Integration by parts is a technique in calculus used to evaluate integrals that are in the form of a product of two functions. It involves rewriting the integral in a different form, using the product rule from differentiation, and then solving for the new integral. This method is particularly useful for integrals involving polynomials, exponential functions, and trigonometric functions.

2. When should I use integration by parts?

Integration by parts should be used when the integrand (the function being integrated) is a product of two functions that cannot be easily integrated using other methods, such as substitution or trigonometric identities. It is also useful when the integrand contains a polynomial and an exponential or logarithmic function.

3. What is the formula for integration by parts?

The formula for integration by parts is ∫udv = uv - ∫vdu, where u and v are two functions and du and dv are their respective differentials. This formula is derived from the product rule for differentiation, d(uv)/dx = u(dv/dx) + v(du/dx).

4. How do I choose u and dv when using integration by parts?

When using integration by parts, the choice of u and dv can greatly affect the ease of solving the integral. A general rule of thumb is to choose u as the function that becomes simpler when differentiated, and dv as the function that becomes simpler when integrated. Some common choices for u include logarithmic functions, inverse trigonometric functions, and polynomials, while common choices for dv include exponential functions, trigonometric functions, and functions involving radicals.

5. Are there any special cases or exceptions when using integration by parts?

Yes, there are a few special cases to keep in mind when using integration by parts. For example, if the integral involves a repeated factor, such as ∫xsin(x)dx, then the u-substitution method may be more efficient. Additionally, when using integration by parts multiple times, it is important to choose u and dv carefully each time to avoid getting stuck in a loop. Lastly, integration by parts does not work for definite integrals involving discontinuous functions, such as step functions.

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