# Finding the general integral

1. Sep 6, 2006

### BlackMamba

Hello,

I have a problem that I am having difficulties with. I'm told to find the general integral.

So here is the problem:
$\int_{\frac {\pi} {2}}^{\frac {-\pi} {2}} \frac {sinx} {1 - sin^2 x} dx$

Here is my partial solution:

$\int_{\frac {\pi} {2}}^{\frac {-\pi} {2}} \frac {sinx} {cos^2 x} dx$

$\int_{\frac {\pi} {2}}^{\frac {-\pi} {2}} \frac {1} {cosx} * \frac {sinx} {cosx} dx$

$\int_{\frac {\pi} {2}}^{\frac {-\pi} {2}} secxtanx dx$

$= [tanx]$

But here is where I am stuck, because when you evaluate $tanx$ from $\frac {\pi} {2}$ to $\frac {-\pi} {2}$ you get an error because no value exists for $tan \frac {\pi} {2}$

Any help would be greatly appreciated.

2. Sep 6, 2006

### fourier jr

wouldn't it be zero? you're integrating from pi/2 to -pi/2 & tan is an odd function.

3. Sep 6, 2006

### BlackMamba

Well that is what I was unsure of. I just double checked the math on my calculator, and I realized I made an error inputting the values. So that would explain the error.

Thanks for the input.

4. Sep 6, 2006

### d_leet

The problem might be that you have the wrong antiderivative of sec(x)tan(x). What is the derivative of sec(x)?

5. Sep 7, 2006

### fourier jr

OOPS i guess it has been a while since i last did calculus... sorry about that

6. Sep 7, 2006

### Tomsk

But surely sec (pi/2)=1/0 as well??:yuck:

7. Sep 7, 2006

### HallsofIvy

Staff Emeritus
I'm a little puzzled by your saying you are asked to find the "general integral" and then asking for a specific value for a definite integral.

Of course, since the denominator, 1- sin2x is 0 at both $\frac{pi}{2}$ and $-\frac{\pi}{2}$, this is an "improper integral". It's value is defined as
$$\lim_{\alpha\rightarrow -\frac{\pi}{2}} \lim_{\beta\rightarrow \frac{\pi}{2}}\int_\beta^\alpha \frac{sin x}{1- sin^2 x}dx$$
So what you want to do is find the "general anti-derivative" and then take limits. Of course, that is precisely what you did- you can't evaluate tan(x) at the endpoints precisely because those limits do not exist.

The "Cauchy Principal Value" for this integral is defined as
$$\lim_{\alpha\rightarrow \frac{pi}{2}}\int_\alpha^{-\alpha}\frac{sin x}{1- sin^2x} dx$$
and since the integrand is odd, that is 0.

Last edited: Sep 7, 2006