# Finding the general integral

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

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

OOPS i guess it has been a while since i last did calculus... sorry about that  d_leet said:
The problem might be that you have the wrong antiderivative of sec(x)tan(x). What is the derivative of sec(x)?

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

HallsofIvy
Homework Helper
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.

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