What method would I use to integrate this?
sqrt(4x^2+25x^4)

Question

anonymous · Answer

I'm thinking trig sub but I can't figure out how to weasel that in.

dumbcow · Answer

$$x = \frac{4}{3} \tan u$$
$$dx = \frac{4}{3} \sec^2 u du$$

tkhunny · Answer

Can you assume x > 0?

$\sqrt{x^{4} + 16x^{2}} = |x|\sqrt{x^{2} + 16}$

And a much simpler substitution may be possible.

anonymous · Answer

so your equation is$$\sqrt{9x^4+16x^2}$$you can pull out a common term of x^2 from both terms, and get$$ \pm x \sqrt{9x^2+16}$$integrate that and you're good

tkhunny · Answer

One cannot integrate "$\pm$".  Use "+" for $x\ge 0$ and "-" for $x < 0$.

anonymous · Answer

I do believe that the x is positive, between 0 and 1. I see what you're saying. So I can take the x^2 out of the radical completely that way and then substitute tangeant inside it?

tkhunny · Answer

Tangent?  I guess if you really want.  I'd try  u^2 = 9x^2 + 16

dumbcow · Answer

yeah do that ^^

anonymous · Answer

I'm trying to do that but I'm usually awful at visualizing u substitution for some reason. :P So u^2 = 9x^2+16. Is du=18x? So you get (sqrt(u) du)/18?

dumbcow · Answer

you need to differentiate the u^2 as well

--> 2u du = 18x dx

--> dx = u/9x dx

anonymous · Answer

Oh wow and that's it? All that's left to do is plug stuff back in to the u and simplify?

dumbcow · Answer

yeah after integrating

anonymous · Answer

Yeah I just realized that as I was writing. So to integrate that do you need to put 9x into terms of u?

dumbcow · Answer

normally yes, but in this case it will cancel out with the "x" you factored out of sqrt

anonymous · Answer

I'm pretty confused here. Where does the extra t come in to cancel?

tkhunny · Answer

"t"?

anonymous · Answer

I'm sorry I meant x

tkhunny · Answer

Simplify your life a little.

$u = 9x^{2} + 16$

$du = 18x\;dx$

$\int x\sqrt{9x^{2} + 16}\;dx = \dfrac{1}{18}\int 18x\sqrt{9x^{2} + 16}\;dx = \dfrac{1}{18}\int \sqrt{u}\;du = \dfrac{1}{18}\dfrac{u^{3/2}}{3/2} + C$

anonymous · Answer

As I said before these substitutions kill me. XP That makes a bit more sense to me. I'll try to roll through it that way and see what happens. Turns out I thought I got the correct answer before but I didn't.

anonymous · Answer

so in the end you get:
$$\frac{ 1 }{ 27 }(9t^2+16)^{3/2}$$
right?

anonymous · Answer

Yes that seems to have done it for sure this time. Thanks a lot! I wish I could give medals to all of you!

tkhunny · Answer

Except for your missing Constant.

You can always find the 1st derivative and check!