Hello, I am absolutely stumped on this one. I'm assuming all of my work is correct up until the end, but I don't know how to proceed using Trigonometric Substitution. Please help!
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Originally Posted by Troopa Hello, I am absolutely stumped on this one. I'm assuming all of my work is correct up until the end, but I don't know how to proceed using Trigonometric Substitution. Please help! $\displaystyle \frac 1{\sec^2 \theta} = \cos^2 \theta = \frac 12 (1 + \cos 2 \theta)$ (and of course it's not $\displaystyle \frac 1{\tan \theta + C}$ there is no integration rule to justify such a step!)
So then... $\displaystyle = \frac 12 (1 + \cos 2 \theta)$ $\displaystyle = \frac 12 (\theta + (cos2 \theta^2 / 2))$ Is this right?
Originally Posted by Troopa So then... $\displaystyle = \frac 12 (1 + \cos 2 \theta)$ $\displaystyle = \frac 12 (\theta + (cos2 \theta^2 / 2))$ Is this right? no. first, where's your integral sign? secondly, how do we integrate cosine? We don't just simply integrate the argument.
$\displaystyle \int\frac{1}{\sec^2\theta}d\theta=\int\cos^2\theta {d}\theta=\frac{1}{2}\int(1+\cos2\theta)d\theta=\f rac{1}{2}(\theta+\frac{1}{2}\sin{2\theta})+C$ Now back substitute.
Thanks for your help. I'm gonna finish this and get some help in my math lab.
Last edited by Troopa; Jan 21st 2010 at 09:27 PM.
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