Question

does anyone have any help integrating this.
(y^2)/(36-(y^2))^(3/2)

Answer 1

trig sub

Answer 2

try with y=6sint....you will get trigonometric integral...

Answer 3

I dont know what to do....@Outkast3r09 and @nenadmetamatika

Answer 4

y=6sin(t)=asin(t)
dy=6cos(t)dy=acos(t)dy
\[\sqrt{36-y^2}=\sqrt{36-36sin^2t}=\sqrt{36(1-sin^2(t)} = 6 \sqrt{1-sin^2t}=\[6\sqrt{cos^2t}\]

Answer 5

\[=6\sqrt{cos^2t}=6cos(t)\]

Answer 6

replace these all in your equation and you'll have integral that's much easier

Answer 7

thanks

Answer 8

should get

\[\int{}{}\frac{36sin^2t}{(6cos(t)^{3/2}}*6cos(t)dt\]

Answer 9

thats not one of the answers i have. there were four choices given.

Answer 10

what are the answers

Answer 11

not to know you can simplify it

Answer 12

a. square root (36-y^2)-sin^-1 (y/6) +c
b. 6y/sqrt(36-y^2)-sin^-1 (y) +c
c. y/sqrt(36-y^2)+c
d. y/sqrt (36-y^2)-sin^-1 (y/6) +c

Answer 13

so ive used this website but i just dont understand it at all. http://www.wolframalpha.com/input/?i=y%5E2%2F(36-y%5E2)%5E(3%2F2)%20

Answer 14

get rid of the 3/2 it should just be this

Answer 15

why do you get rid of it?

Answer 16

\[\int{}{}\frac{36sin^2t}{6cos(t)^3}*6cos(t)dt \]

Answer 17

so the answer is the last one?

Answer 18

yes

you end up after simplifying

\[\int{}{}\frac{36sin^2t}{36cos^2t}dt\]

Answer 19

which is the same as

\[36\int{}{}tan^2tdt\]

Answer 20

and you have to draw a triangle to figure out the tangent?

Answer 21

which means tangent is equal to y/(squareroot (36-y^2))

Answer 22

and i meant 36/36 which is one

\[\int{}{}tan^2tdt\]

Answer 23

do you know how to solve this equation?

Answer 24

the equation would be (sec^2 t)-1 then integrate?

Answer 25

yes

Answer 26

making it tan t-t +c

Answer 27

yep then take you trig sub

y=6sin(t)

and solve for t

Answer 28

plug in y/(sqrt 36-y^2) - sin^(-1) (y/6) +c

Answer 29

yep

Answer 30

sweet thanks!!