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Reduction formula problem

  • Thread starter lionely
  • Start date
  • #1
576
2

Homework Statement





Homework Equations





The Attempt at a Solution



Here is my attempt :

19axlf.jpg



I just can't get the answer the writing near the bottom is me trying to get the left hand side , so you can basically ignore that. But I don't think I made the correct first step... any advice?

Didn't notice the picture was so blurry, the first thing I did was just proceed to integrate In by parts.
 

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Answers and Replies

  • #2
BiGyElLoWhAt
Gold Member
1,560
113
Yea your parts wasn't done properly, check it:

##\int_{0}^{\frac{\pi}{2}} e^{2x}sin^nx ##
gotta do a parts:
##u = e^{2x}##
##dv = sin^nxdx##
##du=2e^{2x}##
##v = ...##
gotta do another parts and use ##sin^2 + cos^2 =1##

##\int_{a}^bsin^nxdx = \int_{a}^b[sin^{n-2}(x)][1-cos^2(x)]dx =
\int_{a}^b sin^{n-2}(x)-\int_{a}^{b} sin^{n-2}(x)cos(x)cos(x)## ...
this just ends up at the reduction formula, and I don't feel like going all the way there, 2 more parts then nest them back together, but the point is: once you get your v
##u = e^{2x}##
##dv = sin^nxdx##
##du=2e^{2x}##
##v = ...##
plug it back into the first set of parts and you'll get something really messy that you need to tidy up. I bet that'll simplify to what you have going on in your homework problem.

PS I can't really read the picture with your work
 
Last edited:
  • #3
576
2
I did not think of separating it then making subbing sin^2 x . Thanks.. I'm such an idiot. I just proceeded to integrate by parts without calling sin^n x , sin^(n-2) x* sin^2x
 
  • #4
576
2
I still don't get you fully, here is my working :
yyli8.jpg


Ignore the first line. where I said In = e^2x sinx^(n-1) * sin^x

I also can't really do parts by the u, dv way or whatever. My teacher said he didn't want me to learn it like that. I just know I should always try to integrate the easier one first and differentiate the other one..
 
Last edited:
  • #5
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,236
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I still don't get you fully, here is my working :
[ IMG]http://i59.tinypic.com/yyli8.jpg[/PLAIN]

Ignore the first line. where I said In = e^2x sinx^(n-1) * sin^x

I also can't really do parts by the u, dv way or whatever. My teacher said he didn't want me to learn it like that. I just know I should always try to integrate the easier one first and differentiate the other one..
What do you mean by
"I also can't really do parts by the u, dv way or whatever." ?


For the second integration it looks like you have:
##u=\sin^{(n-1)}(x)\,\cos(x)\ ## and ##\ dv=e^{2x}dx \ .##​
Those are good choices.

What is the derivative of
##u=\sin^{(n-1)}(x)\,\cos(x)\ \ ?##​

It's not what you have.
 
  • #6
576
2
I mean when I carry out integration by parts, I don't call things u or dv or v. I just proceed to integrate. That is how I was taught. The derivative of that would be (n-1)sin^(n-2)xcos^2x -sin^n
 
  • #7
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
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I mean when I carry out integration by parts, I don't call things u or dv or v. I just proceed to integrate. That is how I was taught. The derivative of that would be (n-1)sin^(n-2)xcos^2x -sin^n
A few parentheses would help here making that derivative:
(n-1)sin(n-2)(x)cos2(x) -sinn(x) .

Notice that cos2(x) = 1 - sin2(x) .

And as you say "integrate the easy part": [itex]\displaystyle \int e^{2x}dx[/itex]

Integration by parts for the integral on your second line:
[itex]\displaystyle \int e^{2x}\sin^{n-1}x \cdot \cos x\,dx[/itex]​

will give you an integrand with two terms after you simplify.

One of the integrals can be written as In , the other as In-2 .
 
  • #8
576
2
I got it out finally, I made a mistake with brackets at the 2nd parts... -_-
 
  • #9
576
2
Thank you guys.
 

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