Question

If I am evaluating the integral and I find x=3sec(theta) and there is an x^3 on top of the integral. Does this become 3sec^3(theta) or 9sec^3(theta)?

Answer 1

anyone?

Answer 2

whats the original integral?

Answer 3

One second

Answer 4

x=(3sec(t)) then x^3 = (3sec(t))^3

Answer 5

\[\int\limits_{}^{}x^3 / \sqrt{x^2+9}\]

Answer 6

x = 3tan(t) then

Answer 7

tan^2 + 1 = sec^2

Answer 8

so do let x=3tan(X) so I get dx = 3sec^2XdX right?

Answer 9

x = 3tan
dx = 3sec^2 dt

x^2 = 9tan^3

Answer 10

So the top is (3tanX)^3? and the bottom is 9(sex^2) then mutlipled by 3sex^2XdX?

Answer 11

id try to keep the sex out of it :)

Answer 12

ha good point.

Answer 13

So it reduces a lot? Will it be 27/9 so be ((3 tan^3(x))/(sec^2X))3sec^2XdX

Answer 14

\[\int \frac{9tan^3(t)}{3sec(t)}3sec^2(t)dt\]
\[\int 9tan^3(t)sec(t)dt\]

Answer 15

so the answer is 9tan^3X?

Answer 16

I think the other sec(t) goes away so its just \[\int\limits_{}^{}9\tan^3(t)dt\]

Answer 17

might have to rename tan^3 into (tan^2)(tan); or simply (sec^2-1)(tan)

Answer 18

Yeah I think I need to do the first? not really sure never done it ha. Professor likes to give us things we havent done before

Answer 19

the other sec doesnt just "go away" :)
\[9\int (sec^2(t)-1)(sec(t)tan(t))dt\]
\[9\int sec^2(t)(sec(t)(tan(t)))-sec(t)tan(t)\ dt\]
\[9sec^3(t)-9\int sec(t)tan(t)\ dt\]
\[\frac{9}{3}sec^3(t)-9sec(t)+C\]

Answer 20

then of course you rename your trig back into its appropriate x parts

Answer 21

Im not sure but I thought the other sec was reduced?

Answer 22

because dx=3sec^2(t) it reduces the 9(sec^2(t))

Answer 23

we came up with this after the trig sub
\[\int \frac{9tan^3(t)}{\cancel{3sec(t)}}\cancel{3}sec^{\cancel{2}}(t)dt\]
\[\int 9tan^3(t)sec(t)dt\]

Answer 24

So I got my trig function wrong? I thought 1+Tan^2(t)=sec^2(t)?

Answer 25

\[\sqrt{(3tan(t))^2+9}\to \sqrt{9tan^2(t)+9}\to\sqrt{9}\sqrt{tan^2(t)+1}\to 3sec^2(t)\]

Answer 26

ugh, typos are pretty lol

Answer 27

\[...\sqrt{9}\sqrt{tan^2(t)+1}\to 3sec(t)\]

Answer 28

you are confusing me.. So I was right?

Answer 29

Ha. So it equals 3sec(t) not 3sec^2(t)?

Answer 30

right; which cancels out the 3 sec(t) from 3sec^2(t); leaving us up top with just: sec(t)

Answer 31

Wait... oh tan^2(t)+1 does equal 3sec^2(t) but we have to take the square root?

Answer 32

I get it now I was just missing the damn square root

Answer 33

yep, trigs a pain lol

Answer 34

thanks