Question

integrate the following:

Answer 1

\[\int\limits_{}^{}\frac{ x^3 }{ \sqrt{x^8 + 1} } dx\]

Answer 2

Its x^3 in the numerator right?

Answer 3

\[\bf\large \int\limits \frac{ x^3 }{ \sqrt{x^8+1} }\]

Answer 4

let u = x^4. if you're familiar with substitution you should have this

Answer 5

you forgot the dx...
what you need to do is.. substitute x^4=t ==> x^8=t^2
and 4x^3dx=dt..

Answer 6

ik i forgot the dx-_- accident :(

Answer 7

hahahah, thanks guys, my friend wanted to confirm this answer, so i just posted it here to check :)

Answer 8

:P, Jhanny

Answer 9

Cheers, Tanvidais. :)

Answer 10

ok wait hold on.

Answer 11

just a question... you have \[\bf\large \int\limits \frac{ x^3 }{ \sqrt{(x^4)^{2}+1} }\] so cant you write the square root part as tan inverse?

Answer 12

integrate this one too??:
\[\int\limits_{}^{}\frac{ 1 }{ \sqrt{sinx} }dx\]

Answer 13

arg, no dx, they need equation editor -_-

Answer 14

sinx= (2tanx/2)/1+tan^2 x/2, i guess? correct me if i ma wrong. i am not too sure if its 1 + .. or 1- ..

Answer 15

I just googled.. it is 1+tan^2 x/2 in the denominator.. hmm. this could help if there wasnt the sqrt.. i am not too sure what next..

Answer 16

Where is THAT identity?? I've never seen \[\large \sin x = \frac{ 2tanx }{ 2 }*\frac{ 1 }{ 1+\tan ^{2}(\frac{ x }{ 2 }) }\]

Answer 17

umm. sin2x = 2tanx/1+tan^2 x.. so just substitute x by x/2..
and its 2tan(x/2).. not 2tanx/2.. sorry if i mistyped earlier.

Answer 18

x_x i'm confused, what did you google.... I just want to see some one work out the identity :) I'd like to learn this :P

Answer 19

http://answers.yahoo.com/question/index?qid=20080529151339AAvBjm1

Answer 20

There you go. :)

Answer 21

and i dont think i can integrate the second one.

Answer 22

I atleast understand the sin 2x identity now:) haha

Answer 23

http://www.wolframalpha.com/input/?i=integration+1%2Fsqrt+sinx
Wolf says this.. if anyone can understand it. :P

Answer 24

what math is this??? lol. elliptical integral of the first kind?

Answer 25

hyperbolic identity?

Answer 26

oh mimi is so intelligent :D

Answer 27

eh we are done with the original question Mimi..
1/sqrt(sinx) is the question. :P

Answer 28

\[\large \color{darkorchid}{sinhx=\frac{ e ^{x}-e ^{-x} }{ 2 }}\]

Answer 29

umm. arent we done with x^3 dx/ sqrt ( x^8 + 1) ??

Answer 30

Yeah. She was just answering both questions

Answer 31

Ohhh sorry. :P

Answer 32

This is for the second one? Looks more like the first one haha. Oh it is the first one...

Answer 33

Lol Mimi.. we were done with the orginal question.. then he asked the second question.. 1/ sqrt sinx.. :P

Answer 34

We're all stuck on \[\large \int\limits \frac{ 1 }{ \color{green}{\sqrt{\sin x}} }dx\]

Answer 35

@.Sam.

Answer 36

Let
u=x^4
du=4x^3dx
du/4=x^3dx
\[=\frac{1}{4} \int\limits \frac{1}{\sqrt{1+u^2}} \, du\]
Then use trig sub since a^2+x^2, we'll use
\[u=\tan \theta \\ \\ du=\sec^2\theta d \theta\]
\[=\frac{1}{4} \int\limits \frac{1}{\sqrt{1+u^2}} \, du\]
\[=\frac{1}{4} \int\limits \frac{1}{\sqrt{1+tan^2 \theta}} sec^2\theta\ d\theta \]
Using \(1+\tan^2 \theta=\sec^2\theta\),
\[=\frac{1}{4} \int\limits \frac{1}{\sqrt{\sec^2\theta}} sec^2\theta\ d\theta \]
\[=\frac{1}{4} \int\limits sec\theta\ d\theta \]

Answer 37

SAM SAM SAM.. WE ARE DONE WITH THAT QUESTION. :P
WE ARE STUCK ON A DIFFERENT QUESTION.. 1/sqrt (sinx) ..
he asked that later in the thread..

Answer 38

ew

Answer 39

:P