Question

Integration question. I will type it nicely below in comments.

Answer 1

\[\int\limits_{}^{}\frac{ x ^{3} }{ \sqrt{x ^{2}+81} }dx\]

Answer 2

Do I make
\[x=9\tan \theta \]?

Answer 3

Wait some minutes, I will think about it and write a tip for you!

Answer 4

Sounds good! I've worked through the problem a few times now, but somewhere I'm making a mistake..

Answer 5

try u =x^2 +81

Answer 6

But the derivative of that is x... I don't have an x-term to replace for du. the numerator is cubed.

Answer 7

Ok, here I am! So, do you know anything about integration by parts?

Answer 8

Yes, I know how to do that.

Answer 9

du =2xdx
x^2 =u -81
(u-81)*du =x^2*2xdx

Answer 10

Because the hw specifies to use trig substitution where x=9tan(theta), but if there's another method I'd be willing to go by that.

Answer 11

Try to do this \(\frac{ x ^{3} }{ \sqrt{x ^{2}+81}}=x^2\cdot\frac{ x }{ \sqrt{x ^{2}+81}} \) and integrate by parts!

Answer 12

Alright. I'll try it that way instead. I hate this trig sub stuff, it's not even needed half the time...

Answer 13

You have no need to do a trig substitution!

Answer 14

if you must use trig then x=9tan t is right
but @Algebraic! 's way is better imo

Answer 15

i would go for sub & integration by parts as well

Answer 16

Can you integrate \(\int \frac{ x }{ \sqrt{x ^{2}+81}} dx\) ?

Answer 17

Well, my method is not graded, but the problem does specify trig sub... however, the answer is all i need, so i'm gonna go with the easiest way to get it done.

Answer 18

Yes, klimenkov, that's u-sub

Answer 19

Please, continue! Derivate x^2 and use the integration by parts formula.

Answer 20

actually trig sub.s turns out to be pretty easy... not as easy as u subs but w/e

Answer 21

i actually feel x=9 tan theta would be much simpler than integration by parts.....but since u are not comfortable with trignometric subs., carry on with int. by parts.

Answer 22

\(\int \frac{ x }{ \sqrt{x ^{2}+81}} dx=\frac12\int \frac{ d(x^2) }{ \sqrt{x ^{2}+81}} =\sqrt{x^2+81}+C\)

Answer 23

@Algebraic! 's way is definitely the fastest, and best in my opinion

Answer 24

nopes fastest and easiest is trig. substitution, that is for sure...

Answer 25

too many cooks...
lol

Answer 26

Yea, well i know how to do trig sub... but i get the integral:
\[\int\limits_{}^{}\tan ^{3}\theta \sec \theta d \theta = \int\limits_{}^{}\tan ^{2}\theta \sec \theta \tan \theta d \theta \] But after I sub (1-sec^(theta)) for tan^2(theta) and get a du of sectan.... idk i don't get the right answer..

Answer 27

sec^2 i meant... 1-sec^2

Answer 28

81* (sin^3 / cos^2) if I'm not mistaken...

Answer 29

because its sec^2 theta -1 not 1-....

Answer 30

t^2-1 dt

Answer 31

\[\int{x^3\over\sqrt{x^2+81}}dx\]\[u=x^2+81\implies x^2=u-81\]\[du=2xdx\implies xdx=\frac12du\]\[\frac12\int{(u-81)du\over\sqrt u}=\frac12\int u^{1/2}-81u^{-1/2}du\]as per @Algebraic!
somebody who can solve it in fewer lines wins my deepest respect :)

Answer 32

Oh... wait, i got u^2-1 du as well though xD i think i just wrote it wrong above... because then i have u^3/3 - u and when i solve my triangle for sec(theta) to sub back in i get it wrong xD

Answer 33

le challenge @hartnn

Answer 34

And thanks Turing and Alg.

Answer 35

lol.... :P sorry, i can't solve that in fewer 'lines' .....

Answer 36

Integration by parts:
\(\int x^2\frac{x}{\sqrt{x^2+81}}dx=x^2\sqrt{x^2+81}-2\int x\sqrt{x^2+81}dx=x^2\sqrt{x^2+81}-\)
\(-\int \sqrt{x^2+81}d(x^2)=x^2\sqrt{x^2+81}-\frac23\sqrt{x^2+81}^3+C\)

Answer 37

So what way is the quickest? :)

Answer 38

this is nice! 3 different ways :D

Answer 39

hahahaha xD this is why math is a boss subject!

Answer 40

And Klim, you are perfectly correct! I looked back at my previous attempt and somehow I got a 3/2 instead of a 2/3, so i must have done some bad arithmetic... at least i was close xD Thanks you everyone for the help!

Answer 41

Ok. Ask again anything interesting!

Answer 42

I guess this is just an incredibly easy integral. Tell your teacher to up his game.

Answer 43

word^ haha :D