Question

Trignometric substitutio of the indefinite integral x^3*sqrt(16+x^2)dx

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

just a sec k?

Answer 3

k

Answer 4

so far this is what i have

Answer 5

kk looks good so far

Answer 6

http://answers.yahoo.com/question/index?qid=20080508091701AAA0Mq0
im fixing to have dinner this link should help you with the rest
im sorry

Answer 7

the answer should be [(16-x^2)^5/2]/5-[16(16+x^2)^3/2]/3+C

Answer 8

you should also put your answer back in terms of x
i might be back on later if you haven't figured it out by then k?

Answer 9

k thanks for your help.

Answer 10

im sorry hai peace

Answer 11

can you please help me?

Answer 12

amistre its really tough

Answer 13

x^3*sqrt(16+x^2) dx; yeah, it can be..

Answer 14

4tan(t) = x^2

Answer 15

ack!!.. 4tan(t) = x

Answer 16

64 tan^3(t) * 4 sec(t) dt ; just gotta determine the dt now

Answer 17

its 4sec^2 theta dtheta

Answer 18

4sec^2(t) dt = dx right?

Answer 19

yea

Answer 20

64 tan^3(t) * 16sec^3(t) dt then right?

Answer 21

yea

Answer 22

i put the answer above for reference, this problem is driving me crzy

Answer 23

1024 (tan(t)sec(t))^3

Answer 24

can you do it on your own and see if you get that?

Answer 25

http://www.wolframalpha.com/input/?i=int%2864+tan^3%28t%29+*+16sec^3%28t%29%29+dt

Answer 26

I aint got the time tonight, libraries closing soon

but wolfram gives steps

Answer 27

http://www.wolframalpha.com/input/?i=int%28x^3+sqrt%2816%2Bx^2%29%29dx

or this for the original

Answer 28

You've got it. You've worked your way to a function that can be integrated.

Answer 29

Integral (sec theta tan theta)^3=sec^3 theta. Now you have to convert back. Good job Amistre. He just had to run and didn't have time to interpret it.