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More Trig Subsitution

  • Thread starter teneleven
  • Start date
  • #1
12
0
Aye gents, I've been working on a problem that doesn't have a solution at the back of the book and would be grateful if you'd compare my answer with yours.

Homework Statement



[tex]\int\frac{t^5}{\sqrt{t^2 + 2}}\ dt[/tex]


Homework Equations



[tex]t = \sqrt{2}\tan\theta[/tex]

[tex]dt = \sqrt{2}\sec^2\theta\ d\theta[/tex]


The Attempt at a Solution



[tex]\int\frac{(\sqrt{2}\tan\theta)^5}{\sqrt{(\sqrt{2}\tan\theta)^2 + 2}}\ \sqrt{2}\sec^2\theta\ d\theta[/tex]

[tex]\int\frac{8\tan^5\theta\sec^2\theta}{\sqrt{2\tan^2\theta + 2}\ d\theta}[/tex]

[tex]8\int\frac{\tan^5\theta\sec^2\theta}{\sqrt{2}\sec\theta}\ d\theta[/tex]

[tex]\frac{8}{\sqrt{2}}\int\tan^5\theta\sec\theta\ d\theta[/tex]

[tex]\frac{8}{\sqrt{2}}\int\frac{\sin^5\theta}{\cos^6\theta}\ d\theta[/tex]

[tex]\frac{8}{\sqrt{2}}\int\frac{[(1 - \cos^2\theta)(1 - \cos^2\theta)]}{\cos^6\theta}\ \sin\theta\ d\theta[/tex]

[tex]\frac{8}{\sqrt{2}}\int[\frac{1}{\cos^6\theta} - \frac{2}{\cos^4\theta} + \frac{1}{\cos^2\theta}]\ \sin\theta\ d\theta[/tex]

[tex]\frac{8}{\sqrt{2}}\ [\frac{-1}{7}\ln([\frac{\sqrt{2}}{\sqrt{t^2 + 2}}]^7) + \frac{2}{5}\ln([\frac{\sqrt{2}}{\sqrt{t^2 + 2}}]^5) - \frac{1}{3}\ln([\frac{\sqrt{2}}{\sqrt{t^2 + 2}}]^3)]\ +\ C[/tex]


Thanks.
 
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Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
Where the heck are all of the logs coming from??? Besides, you don't even need to do a trig substitution for this one. Just put u=t^2+2.
 
  • #3
VietDao29
Homework Helper
1,423
1
As Dick has pointed out, you are all correct from the start, except the last part, why should ln appear in that expression?
It should be continued by letting u = cos x like this:

[tex]...= - \frac{8}{\sqrt{2}} \int \left( \frac{1}{\cos ^ 6 \theta} - \frac{2}{\cos ^ 4 \theta} + \frac{1}{\cos ^ 2 \theta} \right) d(\cos (\theta)) = \frac{8}{\sqrt{2}} \left( \frac{1}{5 \cos ^ 5 \theta} - \frac{2}{3 \cos ^ 3 \theta} + \frac{1}{\cos \theta} \right) + C[/tex]

Now, change everything back to t.

[tex]t = \sqrt{2} \tan \theta , \quad \theta \in \left[ -\frac{\pi}{2} , \ \frac{\pi}{2} \right][/tex]

[tex]\Rightarrow \tan \theta = \frac{t}{\sqrt{2}}[/tex]

Note that [tex]\theta \in \left[ -\frac{\pi}{2} , \ \frac{\pi}{2} \right] \Rightarrow \cos \theta \geq 0[/tex], so:

[tex]\cos \theta = \sqrt{\frac{1}{1 + \tan ^ 2 \theta}} = \sqrt{\frac{2}{2 + t ^ 2}}[/tex]

Just plug all that in and see what you get. :)

-----------------

Or, even faster, you can do it like this:
[tex]\int \frac{t ^ 5}{\sqrt{t ^ 2 + 2}} dt = \frac{1}{2} \int \frac{t ^ 4}{\sqrt{t ^ 2 + 2}} d(t ^ 2 + 2)[/tex]

[tex]= \int t ^ 4 d(\sqrt{t ^ 2 + 2}) = \int \left( (\sqrt{t ^ 2 + 2}) ^ 2 - 2 \right) ^ 2 d(\sqrt{t ^ 2 + 2}) = ...[/tex]
Can you go from here? :)
 
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  • #4
HallsofIvy
Science Advisor
Homework Helper
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934
That looks like the hard way to me. Rewrite the integral as
[itex]\int \frac{t^5}{\sqrt{t^2+ 1}dt= n\int \frac{(t^2)^2}{\sqrt{t^2+1} (dt)[/itex]so, with u= t2+1[/itex]] we have
[tex]\int \frac{(u-1)^2}{u^{a+1)}du[/tex]
 
  • #5
12
0
What's wrong with having natural logs in the solution? How does one know to avoid ln when approaching this problem?

I tried Dick's substitution of [tex]u = t^2 + 2[/tex] and didn't get very far.

[tex]u = t^2 + 2[/tex]

[tex]du = 2t\ dt[/tex]

[tex]\frac{1}{2}\int\frac{t^4}{\sqrt{t^2 + 2}}\ 2t[/tex]

Integration by parts after this step looks like something I want to avoid. Maybe I'm a bit unclear of how to use Dick's substitution?


Thank you for the ideas, VietDao, but I'm thrown off by the suggestion...

[tex]\frac{1}{2} \int \frac{t ^ 4}{\sqrt{t ^ 2 + 2}} d(t ^ 2 + 2)[/tex]

Shouldn't the [tex]d(t^2 + 2)[/tex] read [tex]d(2t)[/tex] ?


HallsofIvy, I'm having trouble understanding how the following is legal using a substitution of [tex]u = t^2 + 1[/tex]:

[tex]\int\frac{(u-1)^2}{u^(^a^+^1^)}du[/tex]


Thanks.
 
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  • #6
Gib Z
Homework Helper
3,346
5
He never said anything was wrong with Natural logs in a solution, maybe he was pointing out that they just werent correct at all.

Dick's suggestion should go more like this.
You have the right set up, but forgot the dt: [tex]\frac{1}{2}\int\frac{t^4}{\sqrt{t^2 + 2}}\ 2t dt[/tex]

Now We make the substitution, u=t^2 + 2, so the denominator is sqrt of u!,

the 2t dt, as you correctly wrote, is du.

t^4 is (u-2)^2.

So you get the pleasure of doing
[tex]\frac{1}{2}\int \frac{(u-2)^2}{\sqrt{u}} du[/tex] instead :D

VietDao's suggestion is the same as dicks, but we just wrote the t^2+2 as u, didnt we :D

I don't understand halls method either..assuming he left the 1/2 outside the integral out, I understand his numerator and differential but not the denominator..I thought that should be [tex]\sqrt{u+1}[/tex]..
 
  • #7
12
0
Why is ln wrong? If I have the following problem...

[tex]\frac{1}{2}\int\frac{\sin\theta}{cos^2\theta}\ d\theta[/tex]

my first inclination is to solve it as follows:

[tex]\frac{1}{2}\ln(cos^2\theta) + C[/tex]


Fill me in.
 
  • #8
Gib Z
Homework Helper
3,346
5
[tex]\int \frac{1}{x} dx = \ln x[/tex]

But [tex]\int \frac{1}{x^2} =-\frac{1}{x}[/tex]

Have you seen the power rule for differentiation? Use the reverse power rule for integration, except when the power is -1, that is the only case when the reverse power rule does not work and you have to use the Natural Log function.
 
  • #9
Gib Z
Homework Helper
3,346
5
In fact, to see why you are so obviously wrong, try differentiating your result, you won't get your integrand, thats for sure.
 

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