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Integral with trig substitution

  • Thread starter paralian
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[SOLVED] Integral with trig substitution

1. Homework Statement

Find [tex]\int(x^3)/\sqrt{x^2-9}[/tex]

2. Homework Equations

Trig substitution. sin^2 + cos^2 =1, and other things that you can figure out from that.

Half angle formula, cos^2[tex]\theta[/tex]=(1+cos(2[tex]\theta[/tex]) )*.5

3. The Attempt at a Solution

Let x=3*sec[tex]\theta[/tex]
so dx=3*sec[tex]\theta[/tex]*tan[tex]\theta[/tex] d[tex]\theta[/tex]

When I substitute that in and simplify it, I got:

27*[tex]\int(sec^4(\theta) d\theta)[/tex]

And I don't know how to integrate that. Half angle formulas aren't seeming to work.

Thanks!
 

Answers and Replies

737
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Or, you can transform some of the secants into tangents:

[tex]27\int \sec^4 \theta d\theta = 27\int (1+\tan^2 \theta)\sec^2 \theta d\theta[/tex]

and you can go from there.
 
14
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Or, you can transform some of the secants into tangents:

[tex]27\int \sec^4 \theta d\theta = 27\int (1+\tan^2 \theta)\sec^2 \theta d\theta[/tex]

and you can go from there.
[tex]27\int (1+\tan^2 \theta)\sec^2 \theta d\theta[/tex]

[tex]27\int \sec^2 \theta d\theta + 27\int \tan^2 \theta\sec^2 \theta d\theta[/tex]

[tex]27*tan\theta + 9\tan^3 \theta[/tex]

Haha sorry I kind of forgot that [tex]\int \sec^2\theta d\theta[/tex] was tan!
 
737
0
You are almost done. Because the problem was originally in terms of x, you will need to transform [itex]\theta[/itex] back into x.
 
dextercioby
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There's no need for trigonometry. Part integration and an immediate substitution will do the trick.

[tex] \int \frac{x^{3}}{\sqrt{x^{2}-9}}dx=\int x^{2}\frac{x}{\sqrt{x^{2}-9}}dx=\allowbreak x^{2}\sqrt{x^{2}-9}-2\int x\sqrt{x^{2}-9}\,dx=\allowbreak x^{2}\sqrt{x^{2}-9}-\frac{2}{3}\left( \sqrt{x^{2}-9}\right) ^{3} [/tex]

up to a constant of integration.
 
tiny-tim
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… just simplify …

Integration by parts not necessary:

[tex] \int \frac{x^{3}}{\sqrt{x^{2}-9}}dx
=\int x\sqrt{x^{2}-9}dx\,+\,\int \frac{9x}{\sqrt{x^{2}-9}}dx
\,=\,\frac{1}{3}(x^{2}-9)^{3/2}\,+\,9\sqrt{x^{2}-9}
\,=\,(\frac{1}{3}x^2\,+\,6)\sqrt{x^{2}-9}\,.[/tex]
 
1,750
1
Polynomial division?
 
14
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You are almost done. Because the problem was originally in terms of x, you will need to transform [itex]\theta[/itex] back into x.
Right...I always forget that!

[tex]x=3*sec \theta \Rightarrow \theta = Sec^-1 (x/3)[/tex]

[tex]\Rightarrow tan\theta = \sqrt{x^2-9}/3[/tex]

[tex]9 \sqrt{x^2-9} + (x^2-9)^\frac{3}{2} /3[/tex]

Shiny! Thanks
 

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