Question

whats the definite integral from pi/6 to pi/4 of -x/(1-x^4)^1/2

Answer 1

You cannot do this by hand because letting u=1-x^4 its derivative would equate -4x^3 dx, which cannot be substituted into the equation.

To find the result, type in the equation into y= on your calculator. Then go 2ND->Calc (top row of buttons) and hit #7 for integration. It will graph the function, and then ask for a lower and upper bound which is obviously the limits you provided.

Answer 2

ok, thanks a lot

Answer 3

did you try a trig substitution? that might work

Answer 4

that's not true, it's a identity
it's the inverse sin identity.

Answer 5

thats what i'm supposed to do, but i'm extremely lost. What do you mean it's the inverse sin identity?

Answer 6

\[-\frac{1}{2}\sin^{-1} (x^2)dx\] from pi/6 to pi/4

Answer 7

you can derive it, but most teachers (even in college) do not require it
most teachers don't even require you to know them
but some want you to memorize them
that's the only way to recognize them
that integral is the derivative of the inverse sin function

Answer 8

Oh god lol I feel like a dolt. Yes it's an inverse trig ID. Sorry

Answer 9

whoa myininaya, did you just do that? lols
props!

Answer 10

i dont think we're there yet. we're supposed to substitute and let u=x^2

Answer 11

yes lol

Answer 12

then how are you going to integrate the u^2 without a u on the outside...

Answer 13

they are not hard to derive. it is harder for me to memorize more stuff so i derive it everytime

Answer 14

? what u?

Answer 15

i dont know, thats why im so confused

Answer 16

myininaya- the hint that came with the question said to let u=x^2

Answer 17

it's going to get really messy...but you can try u - subbing again....

Answer 18

why let u=x^2 when you can let cos( pheta)=x^2

Answer 19

where is the hint you guys are talking about? don't do that way. lol. hints can be gross

Answer 20

LOLOLOLOLL go myininaya

Answer 21

i mean we can try, but I don't see why

Answer 22

its number 2 on the first page

Answer 23

have you done trig substitution before?

Answer 24

nope

Answer 25

ok let me think about then

Answer 26

ok, thanks for helping me with this

Answer 27

i think i understand what your book wants
it want's u sub and trig id
give me a min to type this out

Answer 28

\[-\int\limits_{\pi/6}^{\pi/4}\frac{x}{\sqrt{1-x^4}}dx\]
substitute u = x^2 and du = 2xdx
\[-\frac{1}{2}\int\limits_{}^{}\frac{1}{\sqrt{1-u^2}}du\]

now you use the trig id
\[\int\limits_{}^{}\frac{1}{\sqrt{1-u^2}} = \sin^{-1} (u)\]

so the answer is
\[-\frac{1}{2}\sin^{-1} (u)\]
sub x^2 back in
\[-\frac{1}{2}\sin^{-1} (x^2)\]

Answer 29

i don't see how we can do this without trig identity or trig substitution.
You have to know something about those trig integral stuff to do this problem

Answer 30

right

Answer 31

doing it this way satisfies the hint of using u-sub ^_^

Answer 32

ok, so then what do i do with the limits? f(b)-f(a)?

Answer 33

just plug them into the antiderivative

Answer 34

yup, since, we plugged x^2 back into u
you just have to plug pi/6 and pi/4 into the answer that we got up there
i stopped putting them in i'm sorry, i got lazy

Answer 35

putting them is the easy part

Answer 36

haha dont apologize, thanks for all your help, you guys saved me :)

Answer 37

cheers

good luck

Answer 38

deriving them is fun too so if you don't want to memorize junk learn to derive them

Answer 39

lols the smart one is to learn to derive them, that way, you can never really forget
because if you forget, just derive it again :D

Answer 40

hahaha i'll remember that

Answer 41

lol right. i have to derive alot of stuff because my memory sucks

Answer 42

are you an engineering major?

Answer 43

no im a math major

Answer 44

fun stuff, good luck with everything

Answer 45

i'm computer engineering major

Answer 46

got one right :D

Answer 47

computer engineering is dreamy
im going to start my phd next semester and i picked to do a research in the theory of cryptography

Answer 48

that sounds interesting and hard at the same time, good luck

Answer 49

dreamy? what does that mean?
and i like cryptography, at least i think i do...it's problem solving yes?
i'm a freshman btw lols

Answer 50

lol i think i want to date it lol

Answer 51

lololls
i love programming
which involves a lot of problem solving
i've only taken one programming class in college so far, and it wasn't special... i mean
compared to the rest of my classes, it was awesome, but .. ehh -_-

Answer 52

which class? I'm a Computer Science major, love programming.

Answer 53

i learned matlab
it's basically just a super calculator
there are a lot of stuff that you can do with it
but it's not really a language...

Answer 54

oh matlab is nuts. Used a ton for scientific programming.