# Thread: Simple question - rather embarrassing!

1. ## Simple question - rather embarrassing!

Hi guys,

Found this site through google, looks great

Anyway, I've just finished my maths degree and am training to be a teacher at the moment. My plan is to teach up to further maths at A-level (UK), so as such I'm going over a lot of the stuff I haven't really looked at since my first undergraduate year. I'm rather embarrassingly stumped on a question, and I know the answer is staring me right in the face, if you guys could point me in the right direction it would be great, thanks

Write $\displaystyle (\sec\theta+\tan\theta)(\sec\theta-\tan\theta)$ in it's simplest form. Use it to evaluate

$\displaystyle \int \frac{1}{(\sec\theta + \tan\theta)^2}d\theta$

The first part evaluates to 1, which means I can write

$\displaystyle \int \frac{1}{(\sec\theta + \tan\theta)^2}d\theta=\int\frac{\sec\theta - \tan\theta}{\sec\theta + \tan\theta}d\theta$

but then I find I'm a bit stuck, the next step would be muchly appreciated

Kind Regards

Stonehambey

2. Originally Posted by Stonehambey
Hi guys,

Found this site through google, looks great

Anyway, I've just finished my maths degree and am training to be a teacher at the moment. My plan is to teach up to further maths at A-level (UK), so as such I'm going over a lot of the stuff I haven't really looked at since my first undergraduate year. I'm rather embarrassingly stumped on a question, and I know the answer is staring me right in the face, if you guys could point me in the right direction it would be great, thanks

Write $\displaystyle (\sec\theta+\tan\theta)(\sec\theta-\tan\theta)$ in it's simplest form. Use it to evaluate

$\displaystyle \int \frac{1}{(\sec\theta + \tan\theta)^2}d\theta$

The first part evaluates to 1, which means I can write

$\displaystyle \int \frac{1}{(\sec\theta + \tan\theta)^2}d\theta=\int\frac{\sec\theta - \tan\theta}{\sec\theta + \tan\theta}d\theta$

but then I find I'm a bit stuck, the next step would be muchly appreciated

Kind Regards

Stonehambey
Multiply top and bottom by $\displaystyle {\sec\theta - \tan\theta}$ again

The denominator becomes 1. The to[p becomes something you can integrate after multiplying out and using identities

3. Of course! Thanks!