Simple integral with Trig

1. Jul 10, 2009

djeitnstine

For some reason I cannot see why :

$$\int \sqrt{1+tan^2 \left( \frac{\pi }{4H}} z \right)} dz = \frac{4H}{\pi} sinh^{-1} \left( tan \left[ \frac{\pi z}{4H} \right] \right)$$

I do not understand so because can't the trig identity $$1+tan^2 \theta = sec^2 \theta$$ be used and then the integrand simplifies to:

$$\int sec \left( \frac{\pi z}{4H}} \right) dz$$

why can't it be done? This leads to a whole different answer obviously...

2. Jul 10, 2009

Civilized

No, it doesn't. I just worked out the proof that the result was the same using either method. I encourage you to do the proof yourself, although personally I could not have done it without converting between trigonometric and exponential expressions a few times. If you want I will post the proof.

Last edited: Jul 10, 2009
3. Jul 10, 2009

djeitnstine

Wait never mind. I now know why when I plugged in the numbers it came out different...calculator was not in radians:rofl:

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