What did I do Wrong with this integration by u-sub

  • Thread starter shreddinglicks
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In summary: What you did is "creative", but not something to copy.You should try it. Much simpler and you might learn something useful. What you did is "creative", but not something to copy.In summary, the conversation discusses the integral of x^2(a^2-x^2)^.5, with limits from 0 to a. The attempt at a solution suggests using the double angle formula for sin^2(2theta) to make the integration easier. However, there are some issues with changing the limits and the person asking the question claims to have already solved it in their own way. The conversation ends with a suggestion to try the suggested method for a simpler solution.
  • #1
shreddinglicks
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Homework Statement



∫x^2(a^2-x^2)^.5 limits from 0 to a

Homework Equations



∫x^2(a^2-x^2)^.5 limits from 0 to a

The Attempt at a Solution



It's way to much for me to type in the short amount of time I have so I included a picture of my work. It's neat and easy to read.

the answer should be (PI/16)*a^4
 

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  • #2
To tell you the truth, I didn't read the whole thing. But I suggest where you have$$
a^4\int_0^a \sin^2\theta \cos^2\theta~d\theta$$you write that as$$
a^4\int_0^a \left(\frac {\sin(2\theta}{2}\right )^2~d\theta$$and immediately use the double angle formula for the ##\sin^2(2\theta)##. It will be much easier.
 
  • #3
LCKurtz said:
To tell you the truth, I didn't read the whole thing. But I suggest where you have$$
a^4\int_0^a \sin^2\theta \cos^2\theta~d\theta$$you write that as$$
a^4\int_0^a \left(\frac {\sin(2\theta}{2}\right )^2~d\theta$$and immediately use the double angle formula for the ##\sin^2(2\theta)##. It will be much easier.

Which Identity did you use to get that?
 
  • #4
##\sin(2\theta)=2\sin\theta\cos\theta##
 
  • #5
You did not change the limits when you did the substitution.

$$a^4\int_0^{\pi/2}\sin^2\theta \cos^2\theta\,d\theta=\frac{a^4}{4}\int_0^{\pi/2} \sin^2(2\theta)\,d\theta$$
 
  • #6
Pranav-Arora said:
You did not change the limits when you did the substitution.

$$a^4\int_0^{\pi/2}\sin^2\theta \cos^2\theta\,d\theta=\frac{a^4}{4}\int_0^{\pi/2} \sin^2(2\theta)\,d\theta$$

Why does my a become PI/2?
 
  • #7
shreddinglicks said:
Why does my a become PI/2?

##x=\sin\theta##, at ##x=a##, ##a=a\sin\theta \Rightarrow \theta=\pi/2##.
 
  • #8
LCKurtz said:
##\sin(2\theta)=2\sin\theta\cos\theta##

I understand that. The problem is when I apply it, it just gives me my original integral.
 
  • #9
Pranav-Arora said:
##x=\sin\theta##, at ##x=a##, ##a=a\sin\theta \Rightarrow \theta=\pi/2##.

I don't see how that matters. a/a = 1 and the arcsin(1) = PI/2

It's the same.
 
  • #10
I know it's a lot to read but can someone please look over my entire work? I am very close to the answer except for the numerator.
 
  • #11
LCKurtz said:
##\sin(2\theta)=2\sin\theta\cos\theta##

LCKurtz said:
To tell you the truth, I didn't read the whole thing. But I suggest where you have$$
a^4\int_0^a \sin^2\theta \cos^2\theta~d\theta$$you write that as$$
a^4\int_0^a \left(\frac {\sin(2\theta}{2}\right )^2~d\theta$$and immediately use the double angle formula for the ##\sin^2(2\theta)##. It will be much easier.

shreddinglicks said:
I understand that. The problem is when I apply it, it just gives me my original integral.

Apparently you don't understand it. Use$$
\sin^2(2\theta) = \frac {1-\cos(4\theta)}2$$And, as another poster has noticed and I missed, the upper limit for ##\theta## should be ##\frac \pi 2##.
 
  • #12
LCKurtz said:
Apparently you don't understand it. Use$$
\sin^2(2\theta) = \frac {1-\cos(4\theta)}2$$And, as another poster has noticed and I missed, the upper limit for ##\theta## should be ##\frac \pi 2##.

So, are you going to try it? It's only one simple step left to integrate to the correct answer.
 
  • #13
I did get the answer. I went to sleep last night, where I'm located it was past midnight. I solved it this morning.

I didn't quite do it the way you guys have suggested.
 

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  • #14
shreddinglicks said:
I did get the answer. I went to sleep last night, where I'm located it was past midnight. I solved it this morning.

I didn't quite do it the way you guys have suggested.

You should try it. Much simpler and you might learn something useful.
 

1. What is u-substitution and how does it work?

U-substitution, also known as the substitution method, is a technique used in calculus to simplify and solve integrals. It involves substituting a new variable, u, for a complicated expression within the integral. This allows for the integral to be rewritten in terms of the new variable, making it easier to solve.

2. When should I use u-substitution?

You should use u-substitution when faced with an integral that involves a function within a function, such as the chain rule. This can be identified by looking for a function and its derivative within the integral.

3. How do I know if I have done u-substitution correctly?

You can check if you have done u-substitution correctly by taking the derivative of your substituted function and comparing it to the original function within the integral. If they are equivalent, then you have done u-substitution correctly.

4. What are common mistakes to avoid when using u-substitution?

Common mistakes to avoid when using u-substitution include forgetting to take the derivative of the substituted function, substituting the wrong function, and forgetting to change the limits of integration to match the new variable.

5. Are there any tips for mastering u-substitution?

Some tips for mastering u-substitution include practicing with different types of integrals, understanding the properties of derivatives, and always double-checking your work. It is also helpful to understand the concept behind u-substitution rather than just memorizing the steps.

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