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

- anonymous

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

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- anonymous

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.

- anonymous

ok, thanks a lot

- myininaya

did you try a trig substitution? that might work

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## More answers

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- myininaya

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- anonymous

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

- anonymous

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

- myininaya

yes lol

- anonymous

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

- myininaya

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

- myininaya

? what u?

- anonymous

i dont know, thats why im so confused

- anonymous

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

- anonymous

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

- myininaya

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

- myininaya

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

- anonymous

LOLOLOLOLL go myininaya

- myininaya

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

- anonymous

its number 2 on the first page

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- myininaya

have you done trig substitution before?

- anonymous

nope

- myininaya

ok let me think about then

- anonymous

ok, thanks for helping me with this

- anonymous

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

- anonymous

\[-\int\limits_{\pi/6}^{\pi/4}\frac{x}{\sqrt{1-x^4}}dx\]
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
\[-\frac{1}{2}\sin^{-1} (x^2)\]

- myininaya

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

- myininaya

right

- anonymous

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

- anonymous

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

- myininaya

just plug them into the antiderivative

- anonymous

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

- myininaya

putting them is the easy part

- anonymous

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

- anonymous

cheers
good luck

- myininaya

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

- anonymous

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

- anonymous

hahaha i'll remember that

- myininaya

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

- anonymous

are you an engineering major?

- myininaya

no im a math major

- anonymous

fun stuff, good luck with everything

- anonymous

i'm computer engineering major

- anonymous

got one right :D

- myininaya

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

- anonymous

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

- anonymous

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

- myininaya

lol i think i want to date it lol

- anonymous

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 -_-

- anonymous

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

- anonymous

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...

- anonymous

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

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