integral sqrt(4t^2 +16t^6) dt from 0 to 2
i got upto integral sec^3 x
rewrite it in simpler form is my first thought
\[\sqrt{4t^2(1+4t^4)}=2t\sqrt{1+(2t)^2}\] maybe?
this can be integrated by parts if i see it right
careful A, your equation above isn't quite right.
\begin{array}c &&\sqrt{1+(2t^2)^2)}\\ +&2t&\int^1 v\\ -&2&\int^2v\\ +&0&--- \end{array} yep, maybe a better setup, but im not sure if this is the right track to go down yet
\[ 2t \sqrt{1+4t^4} \] Now substitute \( u = t^2 \) or \( u = 2t^2 \) if you prefer.
I would use a substitution 2t^2=tan(u).
most integrals are unsolvable, so I wouldnt worry to much about it ;)
it is just very lengthy
We rewrite the expression as \[I=\int\limits_{}^{}2t \sqrt{1+(2t^2)^2}dt\] By substituting \(2t^2=\tan{u} \implies 4tdt=\sec^2(u)du \implies dt=\frac{\sec^2u}{4t}du\), we get: \[I={1 \over 2}\int\limits \sqrt{1+\tan^2u}\sec^2{u}du={1 \over 2}\int\limits \sec^3{u}du.\]
i did integration by parts and got 1/2(secxtanx +ln(sec+tanx) from 0 to 8
@ Mr. Math, yes i did that and got integral 1/2 sec^3 x
i was looking for a way to do it in less steps.
Hmm, I don't have such a way right now, sorry :D
less steps? ... cant see it in lesser
By the way, the integration of [sec(u)]^3 is not an easy one. You will need to use integration by parts or some other technique.
hm. i hope i dont get this on exam
thank you everyone for helping though:)
You can do it if you get it on your exam!
Couldn't agree more with Mr M. Work this problem through. It's a great exercise. And just because it requires more steps than usual should be thought of as a challenge to master rather than an impossibility. I remember when I was at your stage I'd take harder integrals like this and redo them once or twice a day for several days until I had it down cold. Sometimes I wish I still was that kid. ;-)
Lol, you sounded like an old man :P
Either that or I'm still a bit hung over from last night and remembering when that didn't happen.
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