Simple? Integration Problem (Trig Sub?)

[PitchBlack]

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

\intx^{2}/\sqrt{1-x^{2}}

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

 

[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??

 

[PitchBlack]

No its not a homework problem...its a conceptual problem ...but i don't get it, and I'm not that great with calculus to tell you the truth I am 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?

 

[sutupidmath]

No its not a homework problem...its a conceptual problem ...but i don't get it, and I'm not that great with calculus to tell you the truth I am 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?

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?