# Simple? Integration Problem (Trig Sub?)

1. Feb 4, 2008

### PitchBlack

I can't get it! I'm pretty sure it's trig substition

$$\int$$$$x^{2}/\sqrt{1-x^{2}}$$

Its a practice problem, if someone could show me the light (or steps) that would be wonderful

2. Feb 4, 2008

### sutupidmath

well, yeah a trig substitution would work, try to let x=sin(t) , so you will get dx=cos(t)dt
after you substitute it back you willl end up with something like this

integ of (sin(t))^2dt

Is this a homework problem by the way???

Can you go from here, anyway??

3. Feb 4, 2008

### PitchBlack

No its not a homework problem....its a conceptual problem ....but i dont get it, and I'm not that great with calculus to tell you the truth im not a math major i just want to get it! So is this what you mean....

$$\int(sinx)^{2}$$$$/$$$$\sqrt{1-sinx^{2}}$$

so using u subtitution ( or whatever letter you use)...
u=sinx
du= cosx
and since there is no cos in the original then 1/cos(du)

(1/cos)$$\int du(u)^{2}$$$$/$$$$\sqrt{1-u^{2}}$$

and go from there? did i do it right?

4. Feb 4, 2008

### sutupidmath

Well You did not get it right, to be honest. Look, $$\int\frac{x^{2}}{\sqrt{1-x^{2}}}dx$$ now let sin(t)=x, from here after defferentiating we get cos(t)dt=dx, now let us substitute this back to the integral, so the integral will take this form:
$$\int\frac{(sin(t))^{2}}{\sqrt{1-(sin(t))^{2}}}cos(t)dt$$, now remember that (sin(t))^2= 1-(cos(t))^2, so afer we substitute the integral becomes:
$$\int\frac{(sin(t))^{2}}{\sqrt{(cos(t))^{2}}}cos(t)dt$$= $$\int\frac{(sin(t))^{2}}{cos(t)}cos(t)dt$$= $$\int (sin(t))^{2}dt$$, now do u know how to evaluate this one?

Last edited: Feb 4, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook