Question

I'm practicing another integral...

Answer 1

\[\int\limits_{ }^{ } \frac{x^3}{\sqrt{16-x^2}^{\color{white}{\frac{1}{2}}}}dx\]

Answer 2

I am setting u=x^2,
and the x^3 will be split into another u, and the x dx= 1/2 du

Answer 3

\[\frac{1}{2}\int\limits_{ }^{ } \frac{u}{\sqrt{16-u}^{\color{white}{\frac{1}{2}}}}du\]

Answer 4

then, v=16-u
(which means that v=16-x^2)
and, dv=du

Answer 5

\[\frac{1}{2}\int\limits_{ }^{ } \frac{16-v}{\sqrt{v}^{\color{white}{\frac{1}{2}}}}dv\]

Answer 6

so, then I get:
\[\frac{1}{2}\int\limits_{ }^{ } 16v^{-1/2}-v^{1/2}~dv\]\[16v^{1/2}-\frac{1}{3}v^{1/2}+C\]\[16x-\frac{1}{3}x^3+C\]

Answer 7

my answer in terms of v is a bit wrong. I wrote 1/3v^1/2, but it should be
1/3v^3/2

Answer 8

you could have also combine those subs
\[u=\sqrt{16-x^2} \\ u^2=16-x^2 \\ 2u du=-2x dx \\ u du=-x dx \\ \text{ so you would have } \\ \int\limits_{}^{}\frac{x^2 x}{\sqrt{16-x^2}} dx=\int\limits_{}^{} \frac{(16-u^2)(- u du)}{u}\]
I don't think I made mistake but let me know if I did
also did you replace v with
the u's will cancel

Answer 9

oops I didn't finish my one sentence

Answer 10

a bit hard substitution for me...

Answer 11

let me go through yours
but I think you didn't replace v with 16-x^2

Answer 12

I think the way I did it is fairly simple... without any complex substitution taking bite by bite.

Answer 13

ohh yes I didn't

Answer 14

\[16v^{1/2}-\frac{1}{3}v^{3/2}+C\]\[16\sqrt{16-x^2}-\frac{1}{3}(16-x^2)\sqrt{16-x^2}+C\]

Answer 15

I have to go in 10 minutes.

Answer 16

\[\text{ first sub you did } u=x^2 \text{ so } du=2x dx \\ \text{ so we had } \int\limits_{}^{}\frac{x^3}{\sqrt{16-x^2}} dx=\int\limits_{}^{}\frac{u \frac{1}{2} du}{\sqrt{16-u}} \\ \frac{1}{2}\int\limits_{}^{} \frac{u du}{\sqrt{16-u}} \\ \text{ next sub you did was } v=16-u \\ \text{ so we have } dv=-du \\ \frac{1}{2}\int\limits_{}^{} \frac{16-v}{\sqrt{v}} (-dv) =\frac{-1}{2} \int\limits_{}^{}(\frac{16}{\sqrt{v}}-\frac{v}{\sqrt{v}} )dv \\ =\frac{-1}{2}\int\limits_{}^{}(16 v^\frac{-1}{2}-v^\frac{1}{2}) dv\]

Answer 17

oh and you had dv=du

Answer 18

so it should be dv=-du
since v=16-u

Answer 19

you guys confused me all

Answer 20

sorry

Answer 21

let me type something here, will try to get this done.. lemme try

Answer 22

\[\int\limits_{ }^{ } \frac{x^3}{\sqrt{16-x^2}}dx\]I will do my intial method, it is easiest for me

Answer 23

or

Answer 24

also I'm going to type something that I think you know but probably didn't apply correctly
\[\int\limits_{}^{}u^n du=\frac{u^{n+1}}{n+1}+C , \text{ where } n \neq -1 \]

Answer 25

\[u=x^2,~~du=\frac{1}{2}dx~~\]\[\int\limits_{ }^{ } \frac{u}{\sqrt{16-u}}dx\]

Answer 26

hey, I know the powet rule

Answer 27

\[v=16-u~~(=16-x^2),~~~dv=-du~~~~~~~\int\limits_{ }^{ } \frac{u}{\sqrt{16-u}}dx\]

Answer 28

\[\frac{1}{2} \int\limits_{ }^{ } \frac{16-v}{\sqrt{v}}dx\]

Answer 29

the 1/2 is from before, I forgot it in my reply

Answer 30

\[16\sqrt{16-x^2}+\frac{1}{3}(16-x^2)\sqrt{16-x^2}+C\]

Answer 31

\[16\sqrt{16-x^2}+\frac{16}{3}\sqrt{16-x^2}-x^2\sqrt{16-x^2}+C\]\[(16+\frac{16}{3}-x^2)\sqrt{16-x^2}+C\]\[(16+5\frac{1}{3}-x^2)\sqrt{16-x^2}+C\]\[(21\frac{1}{3}-x^2)\sqrt{16-x^2}+C\]

Answer 32

alright I am done