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Integration by trig substitution

  • Thread starter mvantuyl
  • Start date
  • #1
37
0

Homework Statement


Integrate: [tex]\int[/tex][tex]\sqrt{1 - 9t^{2}}[/tex]dt


Homework Equations





The Attempt at a Solution


t = 1/3 sin x
dt/dx = 1/3 cos x
dt = 1/3 cos x dx
3t = sin x

1/3 [tex]\int\sqrt{1 - sin^{2} x}[/tex] cos x dx
1/3 [tex]\int cos ^{2}x dx[/tex]
1/3 [tex]\int(1 + cos 2x) / 2[/tex] dx
1/6 [tex]\int[/tex]1 + cos 2x dx
1/6 (x + 1/2 sin 2x) + C
1/6 (x + 1/2 (sin x cos x)) + C
1/6 (x + 1/2 (sin x [tex]\sqrt{1 - sin ^{2}x}[/tex])) + C
1/6 (x + 1/2 (3t[tex]\sqrt{1 - 9t^{2}}[/tex])) + C

I can't figure out how to get rid of x in the result.
 

Answers and Replies

  • #2
316
0
I think you will have to use the arcsine function, which is the inverse of sine.
Also, there is a small error in your calculations: sin(2x) is not sin(x)cos(x).
 
  • #3
djeitnstine
Gold Member
614
0
t = 1/3 sin x

[tex]x= sin^{-1}3t[/tex]

its just algebra.

Yes [tex]sin2x = 2sin(x)cos(x)[/tex] so the 1/2 goes away
 
  • #4
37
0
This is weird.

The original statement of the problem was:

t = 1/3 sin[tex]\Theta[/tex]
dt/d[tex]\Theta[/tex] = 1/3 cos[tex]\Theta[/tex]
dt = 1/3 cos[tex]\Theta[/tex]d[tex]\Theta[/tex]
3t = sin[tex]\Theta[/tex]

1/3[tex]\int\sqrt{1-sin^{2}}\Theta[/tex] cos[tex]\Theta[/tex]d[tex]\Theta[/tex]
1/3[tex]\int cos^{2}\Theta[/tex]d[tex]\Theta[/tex]
1/3[tex]\int[/tex](1 + cos 2[tex]\Theta[/tex]) / 2 d[tex]\Theta[/tex]
1/6[tex]\int1 + cos2\Theta[/tex] d[tex]\Theta[/tex]
1/6([tex]\Theta + 1/2 sin 2\Theta[/tex]) + C
1/6([tex]\Theta + 1/2(sin\Theta cos\Theta[/tex]) + C
1/6([tex]\Theta + 1/2(sin\Theta\sqrt{1-sin^{2}\Theta}[/tex])) + C
1/6([tex]\Theta[/tex] + 1/2(3t \sqrt{1 - 9t^{2}})) + C

I can't figure out how to get rid of [tex]\Theta[/tex] in the result.

The Bob replied:

Why not using arcsin? Also, there is a factor of two missing in the 3rd step from the end.

The Bob

And I answered:

Perfect! Thanks.

All that is now gone.
 

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