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Integration sqrt(10z - z^2)

  • #1
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trying to integrate [tex] \sqrt{10z-z^2} [/tex]

i have been advised to used trig subsitution wherein i subsitute z = cos theta
however i end up at a dead end

[tex] \int \sqrt{10 \cos \theta - \cos^2 \theta} sin \theta d \theta [/tex]

i could certainly replace the sin

[tex] \int \sqrt{10 \cos \theta - \cos^2 \theta} \sqrt{1- \cos^2 \theta} d \theta [/tex]

hwat now do i expand?
or do i use parts?
 

Answers and Replies

  • #2
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alternatively i can use [tex] z = 10 sin^2 \theta [/tex]

i am currenttly working o ntaht because i made a mistake... z should be cos theta

is that hter ight way to go??
[tex] \int \sqrt{100 sin^2 \theta - 100 \sin^4 \theta} (20 sin \theta cos \theta) dz [/tex]
 
Last edited:
  • #3
GCT
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try [tex]z=10-w[/tex] substitution
 
  • #4
arildno
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The simplest choice is to rewrite your root expression as:
[tex]\sqrt{10z-z^{2}}=5\sqrt{1-(\frac{z-5}{5})^{2}}[/tex]
 
  • #5
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[tex]10z-z^2=5^2-(z-5)^2[/tex]
Try [tex]z-5=5\cos{t}[/tex]
 
  • #6
GCT
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what's more simple than this?

[tex] I = \int \sqrt{10z-z^2} dz [/tex]

[tex]z=10-t,~dz=-dt[/tex]

[tex]I= - \int (10-t) \sqrt{10-(10-t)} dt [/tex]

[tex] I= -10 \int t^{ \frac{1}{2}} dt + \int t^{ \frac{3}{2}} dt [/tex]

arildno, can you explain your solution?
 
  • #7
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GCT said:
[tex] I = \int \sqrt{10z-z^2} dz [/tex]

[tex]z=10-t,~dz=-dt[/tex]

[tex]I= - \int (10-t) \sqrt{10-(10-t)} dt [/tex]
Last line isn't correct. it should be [tex]I= - \int \sqrt{t (10-t)} dt [/tex]
 
  • #8
arildno
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It's the same as Yegor's, since we now make the substitution
[tex]\frac{z-5}{5}=\cos(t)[/tex]
 
  • #9
GCT
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I see, well with yegor's/arildno proposal, I end up with [tex]25 \int cos^{2} \theta d \theta[/tex]
 
  • #10
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GCT said:
I see, well with yegor's/arildno proposal, I end up with [tex]25 \int cos^{2} \theta d \theta[/tex]
.. then just use [itex]\cos 2\theta = 2\cos^{2} \theta - 1[/itex], presumably.
 

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