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Few questions on integration,some i'm afraid I just need a clue as I can't even start

  • Thread starter Sensayshun
  • Start date
  • #1
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Homework Statement



1.)
[tex]\int{\frac{t}{(1+t^{2})^3}dt}[/tex]

2.)
[tex]\int{xe^{-2x}dx.}[/tex]

3.)
[tex]\int{\frac{x.dx}{2+x^{2}}}[/tex]
This one is inegrating between 2 and 0, but I didn't know how to format that in.

4.)
[tex]\int{\frac{cos t}{1 + sin t}dt.}[/tex]


Homework Equations



Q 1.)
I'm substituting [tex]u = 1 + t^{2}[/tex]

Q 2.)
[tex] uv - \int{vdu}[/tex]
where u = x
and
dv = [tex]e^{-2x}[/tex]

Q 3 & 4.)
I'm afraid i'm well and truly stuck with these, I think for number 3 I want to use function over a function rule possibly?



The Attempt at a Solution



Promise not to laugh ok :redface:

1.)
Rearranging to [tex]\int{t.u^{-3}}[/tex]
using [tex] uv - \int{vdu}[/tex]
[tex] t.\frac{-1}{2}u^{-2} - \int{\frac{-1}{2}u^{-2}.t^{2}[/tex]
which is:
[tex] t.\frac{-1}{2}u^{-2} - (u^{-2}.\frac{1}{3}t^{3})[/tex]

Am I anywhere close?

As for the others i'm afraid i've not much idea.

2.) could be

using [tex] uv - \int{vdu}[/tex] again:

[tex]x.-2e^{-2x} - \int{-2e^{-2x}}[/tex]
which goes to
[tex]x.-2e^{-2x} - 4e^{-2x}[/tex]

But once again, I could be wildly wrong.

It's not so much that I think i'm going wrong, it's more that I've no idea how to approach the questions. Any nudges in the right direction would be fantastic

Thank you.
 

Answers and Replies

  • #2
Cyosis
Homework Helper
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For 1) you're not doing your substitution correctly. While the substitution [itex]u=1+t^2[/itex] is a good one, you ignore that this would also mean [itex]du=d(1+t^2)=2tdt \Rightarrow tdt=\frac{1}{2} du[/itex]. Try 1) again now.

Regarding 2). You're pretty close, however instead of integrating the exponential you're differentiating it every time, fix that and you will have solved 2).

3) You will have to do a substitution, what do you think is a good one?

4)Again a substitution, any ideas?
 
  • #3
13
0


Thanks Cyosis. I tried to find a way to give you reputation points but it doesn't seem to exist on this board. Just so you know I really appreciate your help :)

For 3.) and 4.) I would've thought substituting the denominator of each fraction. So:
[tex] u = (2 + x^{2})[/tex] and [tex] u = (1 + sin t)[/tex] respectively.

I'll have a look at 1.) and 2.) now following your help.

p.s. for number 3.) Would it be incorrect to follow the function over a function rule? It's just that I see that the differential of [tex]2 + x^{2}[/tex] would equal [tex]2x[/tex] and seem to remember a way to integrations in the form [tex]\frac{f(x)}{(f'(x)}[/tex] however I could be wrong.
 
  • #4
Cyosis
Homework Helper
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Yep those are the correct substitutions for 3) and 4).

I am not sure what kind of rule you're talking about, but if you've a function where the numerator is the derivative of the denominator you can integrate the expression. This is however just a special case of the 'substitution rule' for integrals.

Example: using the substitution u=f(x), du=f'(x)dx
[tex]
\int \frac{f'(x)}{f(x)}dx=\int \frac{du}{u}=\log u+C=\log f(x)+C
[/tex]
 
Last edited:
  • #5
13
0


So for number 3.)

[tex]\int{\frac{x.dx}{2 + x^{2}}}[/tex]

I could give the answer [tex] 2log(2 + x^{2}) + c[/tex]?

As:

[tex]\int{\frac{x.dx}{2 + x^{2}}} = 2\int{\frac{f'(x)}{f(x)}}}[/tex]
 
  • #6
Cyosis
Homework Helper
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Well not entirely right since the derivative of the denominator needs to be equal to the numerator. In this case [itex]2x \neq x[/itex]. You can however write x as (1/2)2x. Thus you get [itex]\frac{1}{2}\int \frac{2x}{2+x^2}dx=\frac{1}{2} \log{(2+x^2)}+C[/itex].
However I really suggest you do not just use this as a rule, but instead do the substitution yourself so you can practice.

Hint: Always differentiate your expression after integration to see if it's correct.
 
  • #7
13
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Ahhh ok then. I'll work through these and update tomorrow. Thanks for all of your help again.
 
  • #8
13
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Just a thought, I'll work through number 4 with substitution as well. But does that not follow the same rule?

[tex]\int{\frac{cos t}{1 + sin t}dt}[/tex]

Differentiatial of the denominator is [tex]-cos t[/tex]

resulting in:

[tex]-1\int{\frac{cos t}{1 + sin t}dt = -log(1 + sin t) + C?[/tex]

I think I may have gone wrong with the -1 there, not sure.
 
  • #9
Dick
Science Advisor
Homework Helper
26,258
618


Just a thought, I'll work through number 4 with substitution as well. But does that not follow the same rule?

[tex]\int{\frac{cos t}{1 + sin t}dt}[/tex]

Differentiatial of the denominator is [tex]-cos t[/tex]

resulting in:

[tex]-1\int{\frac{cos t}{1 + sin t}dt = -log(1 + sin t) + C?[/tex]

I think I may have gone wrong with the -1 there, not sure.
That would be fine if the derivative of sin(t) were -cos(t), but it's not. It's +cos(t).
 
  • #10
13
0


That would be fine if the derivative of sin(t) were -cos(t), but it's not. It's +cos(t).
Lol. I'm in integration mode, my bad.

Doesn't that just make it even easier?
Because that means that:

[tex]\int{\frac{cost}{1 + sint}dt} = \int{\frac{f'(x)}{f(x)}} [/tex]

therefore:

[tex]\int{\frac{cost}{1 + sint}} = log(1 + sint) + C?[/tex]
 
  • #11
Dick
Science Advisor
Homework Helper
26,258
618


Sure. You can always check yourself by differentiating the result.
 

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