OpenStudy (jtug6):

Another calc 2 problem. Someone mind helping me out?

1 year ago
OpenStudy (jtug6):

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

1 year ago
OpenStudy (jtug6):

er dx on the end haha

1 year ago
OpenStudy (jtug6):

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

1 year ago
OpenStudy (jtug6):

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

1 year ago
OpenStudy (anonymous):

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

1 year ago
OpenStudy (anonymous):

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

1 year ago
OpenStudy (jtug6):

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

1 year ago
OpenStudy (jtug6):

thats wrong isnt it?

1 year ago
OpenStudy (anonymous):

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.

1 year ago
OpenStudy (anonymous):

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

1 year ago
OpenStudy (jtug6):

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

1 year ago
OpenStudy (anonymous):

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

1 year ago
OpenStudy (jtug6):

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

1 year ago
OpenStudy (anonymous):

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

1 year ago
OpenStudy (jtug6):

definitley

1 year ago
ganeshie8 (ganeshie8):

Let $$u = x^{3/2}$$

1 year ago
ganeshie8 (ganeshie8):

$$du = \dfrac{3}{2}x^{3/2-1} \implies \dfrac{2}{3}du = \sqrt{x}dx$$ : $\int \dfrac{\sqrt{x}}{\sqrt{1-x^2}}\,dx = \dfrac{2}{3}\int \dfrac{1}{\sqrt{1-u^2}}\,du = \cdots$

1 year ago
OpenStudy (jtug6):

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

1 year ago
OpenStudy (astrophysics):

inverse trig yeo

1 year ago
ganeshie8 (ganeshie8):

lookup the derivative of arcsin(x)

1 year ago
OpenStudy (astrophysics): 1 year ago
OpenStudy (jtug6):

OH! okay

1 year ago
OpenStudy (anonymous):

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

1 year ago
ganeshie8 (ganeshie8):

$x^3 = (x^{3/2})^2 = (u)^2$

1 year ago
OpenStudy (anonymous):

@ganeshie8 I see now. was just a typo.

1 year ago
ganeshie8 (ganeshie8):

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

1 year ago
ganeshie8 (ganeshie8):

Here I have corrected : $$du = \dfrac{3}{2}x^{3/2-1} \implies \dfrac{2}{3}du = \sqrt{x}dx$$ $\int \dfrac{\sqrt{x}}{\sqrt{1-x^{\color{red}{3}}}}\,dx =\int \dfrac{\sqrt{x}}{\sqrt{1-(x^{\color{red}{3/2}})^2}}\,dx = \dfrac{2}{3}\int \dfrac{1}{\sqrt{1-u^2}}\,du = \cdots$

1 year ago
OpenStudy (jtug6):

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

1 year ago
OpenStudy (anonymous):

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

1 year ago
OpenStudy (jtug6):

right. thanks again

1 year ago