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Mathematics 16 Online
OpenStudy (anonymous):

integral sqrt(4t^2 +16t^6) dt from 0 to 2

OpenStudy (anonymous):

i got upto integral sec^3 x

OpenStudy (amistre64):

rewrite it in simpler form is my first thought

OpenStudy (amistre64):

\[\sqrt{4t^2(1+4t^4)}=2t\sqrt{1+(2t)^2}\] maybe?

OpenStudy (amistre64):

this can be integrated by parts if i see it right

OpenStudy (jamesj):

careful A, your equation above isn't quite right.

OpenStudy (amistre64):

\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

OpenStudy (jamesj):

\[ 2t \sqrt{1+4t^4} \] Now substitute \( u = t^2 \) or \( u = 2t^2 \) if you prefer.

OpenStudy (mr.math):

I would use a substitution 2t^2=tan(u).

OpenStudy (amistre64):

most integrals are unsolvable, so I wouldnt worry to much about it ;)

OpenStudy (anonymous):

it is just very lengthy

OpenStudy (mr.math):

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: \[I={1 \over 2}\int\limits \sqrt{1+\tan^2u}\sec^2{u}du={1 \over 2}\int\limits \sec^3{u}du.\]

OpenStudy (anonymous):

i did integration by parts and got 1/2(secxtanx +ln(sec+tanx) from 0 to 8

OpenStudy (anonymous):

@ Mr. Math, yes i did that and got integral 1/2 sec^3 x

OpenStudy (anonymous):

i was looking for a way to do it in less steps.

OpenStudy (mr.math):

Hmm, I don't have such a way right now, sorry :D

OpenStudy (amistre64):

less steps? ... cant see it in lesser

OpenStudy (mr.math):

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.

OpenStudy (anonymous):

hm. i hope i dont get this on exam

OpenStudy (anonymous):

thank you everyone for helping though:)

OpenStudy (mr.math):

You can do it if you get it on your exam!

OpenStudy (jamesj):

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. ;-)

OpenStudy (mr.math):

Lol, you sounded like an old man :P

OpenStudy (jamesj):

Either that or I'm still a bit hung over from last night and remembering when that didn't happen.

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