- #1
What sum ?uzman1243 said:Homework Statement
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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?
SammyS said:What sum ?
Let u = sin(n-1)(x), and dv = sin(x) dx
(The King beat me by fractions of a minute!)
(n-1) is half way from n to (n-2) .uzman1243 said:Lol. He did.
Anyways, can you tell me why sin(n-1)(x)?
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.
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.
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.
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".
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.