trig substitution integral: integral of sqrt[(1-x^2)/x^2] dx

[rooney]

integral(square root(1-x^2)/x^2) dx

I used the substitution x = sin(?) for the numerator, and then figured that I could use it for the denominator as well. that is, in the numerator I have square root(1-(sin(?))^2) and (sin(?))^2 in the denominator. for the new operator I have cos(?) d?. i did some algebra, and came up with the integral of 1/(tan(?))^2 d?. I don't know how to integrate this, and when I did it on my calculator it gave me a different answer than the original integral that I started with.

 

[Deleted member 4993]

\(\displaystyle \int\sqrt{\frac{1-sin^2\theta}{sin^2\theta}}\cdot cos\theta \,\,d\theta\)

\(\displaystyle \,\,=\,\,\int\frac{cos^2\theta}{sin\theta}\,\,d\theta\)

\(\displaystyle \,\,=\,\,\int\frac{1-sin^2\theta}{sin\theta}\,\,d\theta\)

\(\displaystyle \,\,=\,\,\int (cosec\theta \,\,-\,\, sin\theta)\,\,d\theta\)

Now you have "standard" trigonometric integrals.

 

[rooney]

Thank you very much for the solution. I am not following what is going on in the denominator, though. I do not understand what is happening when it goes from (sin(?))^2 to sin(?). Could someone please elaborate a little for me?

 

[Deleted member 4993]

Thank you very much for the solution. I am not following what is going on in the denominator, though. I do not understand what is happening when it goes from (sin(?))^2 to sin(?). Could someone please elaborate a little for me?

Use paper/pencil and write down my solution.

Do not forget the square-root - and do not forget the 'cos' outside of the square-root.