Integration by parts with trig function insanity

In summary, the conversation is about proving the equation (I2n+1)/(I2n)=(2n+1)/(2n+2) using the integral from 0 to pi/2 of sinnxdx. The attempted solution includes using integration by parts and induction, and references Wallis' infinite product for pi/2. The accuracy of the formula for I2n is also questioned.
  • #1
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Homework Statement


Let In=the integral from 0 to pi/2 of sinnxdx
show that (I2n+1)/(I2n)=(2n+1)/(2n+2)


Homework Equations


integral from 0 to pi/2 of sin2n=(2n-1)pi/4n


The Attempt at a Solution


I can't write down all that I've done because it's just too ridiculous. I've tried lots of forms of just trying to manipulate my equation above, which didn't work. I've tried integration by parts for sin2n+1*sinx but I've ended up with the integral of cos2xsin2nx and I can't find a suitable substitution for that. I've been working on this problem forever and it's killing me. If anyone has any suggestions I would be very appreciative.
 
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  • #2
Have you seen the derivation of Wallis' infinite product for pi/2?
 
  • #3
unfortunately, I have to use this problem to prove that next.
 
  • #4
so you're trying to do the even integral first? think you might have been on the right track how about looking at some induction as a next step...

I2n = [tex]\int[/tex]dt sint2n = [tex]\int[/tex]dt sint2n-1.sint

integrating by parts

I2n = (2n-1)[tex]\int[/tex]dt sint2(n-1)cost2

I2n = (2n-1)[tex]\int[/tex]dt sint2(n-1) - (2n-1)[tex]\int[/tex]dt sint2n

So
2n. I2n = (2n-1).I2(n-1)
 
  • #5
so you're trying to do the even integral? think you might have been on the right track how about looking at some induction as a next step...

so from I2n = [tex]\int[/tex]dt sint2n = [tex]\int[/tex]dt sint2n-1.sint

and integrating by parts like you did, I get to

2n. I2n = (2n-1).I2(n-1)

Also is that formula for I2n in the post correct? i would of thought it should tend zero as n heads towrds infinity
 

1. What is integration by parts?

Integration by parts is a technique used in calculus to find the integral of a product of two functions. It is based on the product rule from differentiation and involves splitting a complicated integral into two simpler integrals.

2. How does integration by parts work?

Integration by parts involves choosing one function to differentiate and another function to integrate. This creates a new integral that may be easier to solve than the original. The integration by parts formula is ∫u dv = uv − ∫v du.

3. What is the purpose of using trigonometric functions in integration by parts?

Trigonometric functions are commonly used in integration by parts because they often appear in integrals that are difficult to solve. By using integration by parts with trigonometric functions, we can simplify the integral and make it easier to solve.

4. Can integration by parts be used with any type of function?

Yes, integration by parts can be used with any type of function as long as it can be integrated. However, it is most useful when one of the functions is difficult to integrate, such as trigonometric functions or logarithmic functions.

5. How do you know when to use integration by parts?

It is not always obvious when to use integration by parts. However, a good rule of thumb is to look for integrals involving products of functions, especially if one of the functions is difficult to integrate. Also, if you have already tried other integration techniques and they have not worked, integration by parts may be a good next step.

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