Question

Evaluate the integral for x>1/2 ?

Answer 1

x^2/sqrt(64x^2-16)

Answer 2

\[x^2/\sqrt(64x^2-16)\]

Answer 3

\[\int\limits \frac{ x^2 }{\sqrt{64x^2-16} }dx\]
\[64x^2-16\ge 0 ,x^2 \ge \frac{ 1 }{4 },\left| x \right|\ge \frac{ 1 }{2 }\]

Answer 4

Is that all? I'm not entirely sure what that's supposed to show?

Answer 5

So does this mean you are doing an improper integral from x=1/2 to x=infinity?

Answer 6

Are you familiar with trigonometric substitution?

Answer 7

That's what I thought, but looking at the graph it looks like it'd diverge to infinity (right terms?)

I tried entering that in webassign and it said that was wrong

Answer 8

So there's no upper bound? It definitely seems to not converge. At the very least, you can still evaluate the integral right?

Answer 9

Yeah, I did a little bit with it, and ended up with the integral of 1/cos^2(u)sin(u) with u = sec^-1(2x)

Which just plugging that in to wolfram ends up looking really ugly, and I doubt its the right answer as everything else on the assignment so far has been like, 1 digit answers.

Answer 10

\[I=\int\limits \frac{ x^2 }{4\sqrt{4x^2-1} }dx\]
put x=1/y
\[dx=\frac{ -1 }{y^2 }dy\]
\[I=\int\limits \frac{ \frac{ 1 }{ y^2 }*\frac{ -1 }{ y^2 }dy }{\frac{ 4 }{y }\sqrt{4-y^2} }\]
\[I=\int\limits \frac{- y~dy }{ 4y^4\sqrt{4-y^2} }\]

Answer 11

That seems unnecessary. @surjithayer

Answer 12

put \[\sqrt{4-y ^2}=t,4-y^2=t^2,y^2=4-t^2,ydy=tdt\]
\[I=\int\limits \frac{ -t~dt }{ 4(4-t^2)^2t }\]

Answer 13

we can use partial fractions further.

Answer 14

\[\int\limits \frac{x^2dx}{\sqrt{64x^2-16}}=\int\limits \frac{x^2dx}{4\sqrt{(2x)^2-1}}\]\[2x=\sec \theta\]\[dx=\frac{1}{2}\sec \theta \tan \theta d \theta\]
\[\frac{1}{32} \int\limits \frac{\sec \theta \tan \theta d \theta }{\sqrt{\sec^2 \theta -1}} =\frac{1}{32} \int\limits \frac{\sec \theta \tan \theta d \theta }{\tan \theta} \]\[\frac{1}{32} \int\limits \sec \theta d \theta \]\[\frac{1}{32} \int\limits \sec \theta \frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta} d \theta\]
\[\frac{1}{32} \ln| \sec \theta + \tan \theta|+C\]

substitute in, remember that

so you have an integral in terms of x again.