Troubling Trig Integral

**kaiser0792** · March 28th 2010, 09:20 PM

I am currently working on an integral that I finally found the answer to after approaching differently, but I would like to know why my first approach failed -- it led to undefined trig functions and/or ln 0. As far as I can tell, by math logic is okay, unless I'm missing something.

$<br /> \int_{0}^{\pi/2}\frac{cost}{1 + sint} dt <br />$

My initial approach was to multiply by $\frac{1 - sin t}{1 - sin t}$ leading to $\int\frac{cost(1 - sin t)}{1 - sin^2 t}$ and then to $\int\frac{cost(1 - sin t)}{cos^2 t}$ and then to $\int\frac{1 - sin t}{cos t}$ . Rewriting, $\int_{0}^{\pi/2}\frac{1}{cos t} dt - \int_{0}^{\pi/2}tan t dt$ . This leads to $\int_{0}^{\pi/2}sec t dt - \int_{0}^{\pi/2}tan t dt$ . But when I attempted to integrate, it failed. Simple u-subst. works u = 1 + sint,
du = cost dt, but why does the other approach fail?

**sa-ri-ga-ma** · March 28th 2010, 09:50 PM

In both the approachs you get the same answer.
Intg(secθ) - Intg(tanθ)
= ln(secθ + tanθ) -ln(secθ)
= ln[(secθ + tanθ/secθ]
= ln[ 1 + sinθ]

**kaiser0792** · March 28th 2010, 10:12 PM

Thread: Troubling Trig Integral

LinkBack

Thread Tools

Display

Troubling Trig Integral

Similar Math Help Forum Discussions

Foundations of mathematics - something is troubling me

Hilbert's Hotel is Troubling Me

trig integral

A troubling limit

A troubling for me...

Search Tags