Integration by Parts Homework: Get Help Now

In summary, the conversation is about using integration by parts for a given sum involving a function I(n). The suggestion is to try to determine the function for discrete values of n and use the substitution u = sin(n-1)(x) and dv = sin(x) dx. The conversation also includes a hint for solving the problem and a question about the choice of u.
  • #1
uzman1243
80
1

Homework Statement


attachment.php?attachmentid=69417&stc=1&d=1399253516.png

Homework Equations



N/A

The Attempt at a Solution



I can't even begin the attempt because I don't know how you could use intergration by parts for this sum in the first place.
Can you help me out?
 

Attachments

  • Untitled.png
    Untitled.png
    3.4 KB · Views: 428
  • Like
Likes 1 person
Physics news on Phys.org
  • #2
First of all, I(n) is not a sum. I is a function which depends on the exponent of the sine function in the integrand.

You might try to determine what I(n) is for some discrete values of n, like n = 2, 3, 4, etc. and see if a pattern emerges.
 
  • Like
Likes 1 person
  • #3
uzman1243 said:

Homework Statement


attachment.php?attachmentid=69417&stc=1&d=1399253516.png

Homework Equations


N/A

The Attempt at a Solution



I can't even begin the attempt because I don't know how you could use intergration by parts for this sum in the first place.
Can you help me out?
What sum ?

Let u = sin(n-1)(x), and dv = sin(x) dx

(The King beat me by fractions of a minute!)
 
  • Like
Likes 1 person
  • #4
hint
$$\int\! \sin^n(x) \, \mathrm{d}x=\int\! \sin^{n-2}(x)\sin^{2}(x) \, \mathrm{d}x=\int\! \sin^{n-2}(x)\big(1-\cos^{2}(x)\big) \, \mathrm{d}x$$
 
  • Like
Likes 1 person
  • #5
This does not look like precalculus.
 
  • #6
SammyS said:
What sum ?

Let u = sin(n-1)(x), and dv = sin(x) dx

(The King beat me by fractions of a minute!)

Lol. He did.

Anyways, can you tell me why sin(n-1)(x)?
 
  • #7
uzman1243 said:
Lol. He did.

Anyways, can you tell me why sin(n-1)(x)?
(n-1) is half way from n to (n-2) .


Try it and see what happens !
 
  • Like
Likes 1 person

1. What is the purpose of Integration by Parts?

Integration by Parts is a technique used in calculus to solve integrals that cannot be solved by basic integration rules. It allows us to break down a complex integral into simpler parts to make it easier to solve.

2. How do I know when to use Integration by Parts?

You should use Integration by Parts when you have an integral that involves a product of two functions, and the integral cannot be solved by any other method. This usually occurs when one function is becoming simpler while the other is becoming more complex.

3. What are the steps for using Integration by Parts?

The steps for Integration by Parts are as follows: (1) Identify the two functions in the integral, (2) choose "u" and "dv" such that the integral of "dv" can be easily solved, (3) use the formula "∫u dv = uv - ∫v du" to solve the integral, (4) repeat the process until the integral is fully solved.

4. Can Integration by Parts be used for definite integrals?

Yes, Integration by Parts can be used for definite integrals. In this case, the formula becomes "∫a to b u dv = [uv]a to b - ∫a to b v du".

5. Are there any common mistakes to avoid when using Integration by Parts?

One common mistake to avoid is choosing "u" and "dv" incorrectly. "u" should be chosen as the function that becomes simpler after each integration, and "dv" should be the function that becomes more complex. Another mistake is forgetting to include the "±C" when integrating "v du" in the formula.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
615
  • Calculus and Beyond Homework Help
Replies
13
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
18
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
24
Views
1K
  • Calculus and Beyond Homework Help
Replies
13
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
Back
Top