Integration using trig substitution

[howsockgothap]

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!

 

[grey_earl]

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 \int \sqrt{9 x^2 - 16} x^{-4} \mathrm{d} x. You substitute x = 4/(3 \cos(y)), so your integral gives \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. Seems you are already wrong here, if I haven't made a mistake, please double check. Of course, then you need
\frac{\mathrm{d}}{\mathrm{d} x} \tan(x) = 1 + \tan^2(x), but I think you can figure it out for yourself from here :)

 

[howsockgothap]

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.

 

[howsockgothap]

Nevermind, figured it out. Thanks.

 

[grey_earl]

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.