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Integration using trig substitution

  • #1

Homework Statement



Integrate (9x2-16)1/2/x4

Homework Equations





The Attempt at a Solution



I set 3x=4secy, 3dx=4secytany and (9x2-16)1/2=2tany.

I then plugged that all into my integral and ended up with2tany*4secytany/(4/3secy)4 dy...
(81/32) tan2y/sec3y dy....

I solved for this by switching tan and sec to cos and sin, so (81/32) cos3y*sin2y/cos2y.... and integrated that to get (81/32)sin3/3+c.... when it's all worked out I had 32/x3

I can't figure out where I went wrong but I'm certain that I haven't got the right answer. Thanks!
 

Answers and Replies

  • #2
169
0
I don't like those sec(x) and cosec(x) functions, I prefer 1/cos(x) and 1/sin(x). But let's get sorted you out.

You have [tex]\int \sqrt{9 x^2 - 16} x^{-4} \mathrm{d} x[/tex]. You substitute [tex] x = 4/(3 \cos(y))[/tex], so your integral gives [tex]\int \sqrt{16/(\cos^2(y)) - 16} \,\cdot 81 \cos^4(y)/256 \,\cdot 4/3\,\cdot 1/(\cos^2(y)) \sin(y) \mathrm{d} y = 4\cdot 81\cdot 4/3/256 \int \sqrt{1-\cos^2(y)} 1/(\cos^2(y)) \sin(y) \mathrm{d} y = 27/16 \int \sin^2(y)/(\cos^2(y)) \mathrm{d} y[/tex]. Seems you are already wrong here, if I haven't made a mistake, please double check. Of course, then you need
[tex]\frac{\mathrm{d}}{\mathrm{d} x} \tan(x) = 1 + \tan^2(x)[/tex], but I think you can figure it out for yourself from here :)
 
  • #3
Not quite sure why you're plugging in 1/cos2ysiny for secytany... shouldn't it simply be 1/cosysiny? I also don't see why you've plugged in (4/3cos2y)*siny. It's supposed to be the dx I realize but the derivative of 4/(3cosy) isn't what you've plugged in, at least as far as I can see. That might be just that I'm not getting it though.

There is one tiny algebra mistake though, you set cos4y/cos2y=1/cos2y, should be cos2y which would give an end result very similar to what I had sin2ycos2y rather than sin2ycosy

I suspect your answer is correct when that little algebra problem is fixed, but I don't understand quite how you got it.
 
  • #4
Nevermind, figured it out. Thanks.
 
  • #5
169
0
Oh, but your sec(y) tan(y) = 1/(cos (y)) sin(y)/cos(y) = sin(y)/cos^2(y), isn't it? And the cos^4(y) I put under the square root. Then, dx = d(4/(3 cos(y))) = 4/3 d(1/(cos(y)) = - 4/3 1/cos^2(y) d(cos(y)) = 4/3 1/cos^2(y) sin(y) dy. Note that two minus signs have cancelled.
 

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