Question

Another calc 2 problem. Someone mind helping me out?

Answer 1

\[\int\limits \sqrt{(x)/(1-x^3)}\]

Answer 2

er dx on the end haha

Answer 3

I was considering first rewriting the 1-x^3 as 1-x^2 * x, and letting x equal sin theta from the form 1-sin^2theta = cos^2 theta

Answer 4

I ended up getting theta plus some constant but I'm pretty sure thats wrong

Answer 5

@jtug6 so you did:
x = sin(u)
\[\int\limits_{}^{}\sqrt{\frac{ x }{ 1-x^{3} }}dx \rightarrow \int\limits_{}^{}\cos(u)\sqrt{\frac{ \sin(u) }{ 1-\sin^3(u) }}du\]

Answer 6

can't use the sin(x)^2 identity in this case however since its sin(x)^3

Answer 7

yes, and i took the 1 - sin^3(theta) or u in this case and made it 1-sin^2(u) * sin(u)

Answer 8

thats wrong isnt it?

Answer 9

What you have written above is correct, but it can't be decomposed any further into the identity above. My first thought is partial fractions, but I still need to think about this one.

Answer 10

so in other words 1-sin(x)^3 is the furthest you can take that

Answer 11

aw. ok. I was going to also say 1-sin(x)^3 was = cos^2(x) * sin(x) then cancel the sin(x) on num/denom but yeah guess thats wrong :p

Answer 12

yeah that doesn't work because that is:
\[\cos^{2}(x)\sin(x)=[1-\sin^{2}(x)]\sin(x)\]

Answer 13

oh, right. distributing the sinx to both. bah my bad. hmmmmm.

Answer 14

i need my whiteboard for this, give me a few minutes to check partial fractions

Answer 15

definitley

Answer 16

Let \(u = x^{3/2}\)

Answer 17

\(du = \dfrac{3}{2}x^{3/2-1} \implies \dfrac{2}{3}du = \sqrt{x}dx\) :

\[\int \dfrac{\sqrt{x}}{\sqrt{1-x^2}}\,dx = \dfrac{2}{3}\int \dfrac{1}{\sqrt{1-u^2}}\,du = \cdots \]

Answer 18

So would we then use a trig rule from 1 - u^2?

Answer 19

inverse trig yeo

Answer 20

lookup the derivative of arcsin(x)

Answer 21

http://tutorial.math.lamar.edu/pdf/Common_Derivatives_Integrals.pdf

Answer 22

OH! okay

Answer 23

The bottom term is 1-x^3 not a quadratic

Answer 24

\[x^3 = (x^{3/2})^2 = (u)^2\]

Answer 25

@ganeshie8 I see now. was just a typo.

Answer 26

Oh right, there is a typo in my previous reply, one sec...

Answer 27

Here I have corrected :

\(du = \dfrac{3}{2}x^{3/2-1} \implies \dfrac{2}{3}du = \sqrt{x}dx \)
\[\int \dfrac{\sqrt{x}}{\sqrt{1-x^{\color{red}{3}}}}\,dx =\int \dfrac{\sqrt{x}}{\sqrt{1-(x^{\color{red}{3/2}})^2}}\,dx = \dfrac{2}{3}\int \dfrac{1}{\sqrt{1-u^2}}\,du = \cdots\]

Answer 28

ahhhh thanks. makes much more sense. was thinking of trig sub as first technique...T_T

Answer 29

jtug6 you can also use the trig rule like you said for the bottom term to find the arcsin function without the table

Answer 30

right. thanks again