Integration by parts MIDTERM really quick question

In summary, the two integrals have the same answer, but the chain rule needs to be applied to get it.
  • #1
banfill_89
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0
integration by parts! MIDTERM...really quick question...please help!

Homework Statement



Evaluate the two following integrals:
[tex]\int[/tex] xcos5x dx and [tex]\int[/tex] ln(2x+1) dx

Homework Equations





The Attempt at a Solution



ok, for the first one, the answer is 1/5 x sin5x + 1/25cos5x+C

i get this answer up until the 1/5x and the 1/25...i have no idea where there coming from. in my solutions manual they come up when u take the integral of cos5x...its 1/5sin5x...where is this 1.5 coming from?

secondly the answer to the second one is: 1/2(2x+1)ln(2x+1)-x+C

again i get this exact answer except for the 1/2. i know its coming from when u take the integral of 1/(2x+1)...but i don't get where the 1/2 comes from in front of the natural log

help would be greatly appreciated...i have a midterm wednesday!
 
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  • #2
chain rule

banfill_89 said:
ok, for the first one, the answer is 1/5 x sin5x + 1/25cos5x+C

i get this answer up until the 1/5x and the 1/25...i have no idea where there coming from. in my solutions manual they come up when u take the integral of cos5x...its 1/5sin5x...where is this 1/5 coming from?

secondly the answer to the second one is: 1/2(2x+1)ln(2x+1)-x+C

again i get this exact answer except for the 1/2. i know its coming from when u take the integral of 1/(2x+1)...but i don't get where the 1/2 comes from in front of the natural log

Hi banfill_89! :smile:

It's the chain rule (backwards!) …

differentiate sin5x and you get an extra 5 because of the chain rule … same with ln(2x+1), you get an extra 2 …

so when you integrate, you have to cancel them out with a 1/5 or 1/2. :wink:
 
  • #3


ohhhhhhh. makes a lot of sense...thanks alot
 

FAQ: Integration by parts MIDTERM really quick question

1. What is integration by parts and how does it work?

Integration by parts is a method used to evaluate integrals of the form ∫u dv. It involves splitting the integral into two parts, choosing one part to differentiate and the other to integrate. The resulting integral can then be solved using integration tables or by using the integration by parts formula.

2. Why is integration by parts important?

Integration by parts is important because it provides a way to evaluate integrals that cannot be solved by other methods such as substitution or partial fractions. It is also useful in solving differential equations and in various applications in physics and engineering.

3. How do I know when to use integration by parts?

There is no definite rule for when to use integration by parts, but it is typically used when the integral involves a product of two functions, one of which can be easily integrated while the other can be differentiated. It is also useful when the integral involves functions with logarithms or inverse trigonometric functions.

4. Are there any limitations to using integration by parts?

Yes, there are some limitations to using integration by parts. It may not always work for every integral, and it can become complicated when used multiple times. It is also important to choose the correct functions to differentiate and integrate in order to avoid getting stuck in an infinite loop.

5. Can you provide an example of integration by parts?

Sure, let's look at the integral ∫x sin(x) dx. We can use integration by parts by setting u = x and dv = sin(x) dx. This gives us du = dx and v = -cos(x). Plugging these into the integration by parts formula, we get ∫x sin(x) dx = -x cos(x) - ∫(-cos(x)) dx. This integral can then be easily solved, giving us the final answer of ∫x sin(x) dx = -x cos(x) + sin(x) + C.

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