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integrate using trig sub. problem below.

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\[\int\limits_{1/2}^{1} dx/ x ^{2}\sqrt{x ^{2} + 4}\]
put x=2tan u
i got as far as \[1/4 \int\limits_{1/2}^{1} \sec ^{2}\theta/ \tan ^{3}\theta+\tan ^{2}\theta \] but im not sure what to do after

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After you plug everything in and factor out the constants, you should have this. \[\large \frac{1}{4}\int\limits \frac{\sec^2\theta \; d\theta}{\tan^2\theta \sqrt{\color{royalblue}{\tan^2\theta+1}}}\] I'm not sure what you did on the bottom.. hmm. From here we want to use an identity, \[\large \color{royalblue}{\tan^2\theta+1=\sec^2\theta}\]Changing the integral to,\[\large \frac{1}{4}\int\limits\limits \frac{\sec^2\theta \; d\theta}{\tan^2\theta \sqrt{\color{royalblue}{\sec^2\theta}}}\]
okay im left with \[1/4\int\limits_{1/2}^{1} \sec \theta/\sec ^{2}\theta -1 \]
No you're left with,\[\large \frac{1}{4}\int\limits \frac{\sec \theta}{\tan^2\theta}d \theta\] You keep making weird substitutions.. I'm not sure why :c Oh you wanted it all in terms of secant I guess? :)
From here, it's probably better to convert everything to sines and cosines and the integral should be fairly simple from there :)
yeah thats what i was trying to do. okay let me try it with sines and cosines
still a bit confused as to what to do with the term thats squared
\[\large \frac{1}{4}\int\limits \frac{\sin \theta}{\cos^2\theta}d \theta\]So we get this I think? Yah it might seem a little tricky at first :) But it's just a simple `U substitution`. Let \(u=\cos \theta\).
Woops did I set that up correctly? I was trying to do the sine and cosine conversion in my head... thinking.
Yah I did that upside down :) woops lol
i think its cos over sin squared
yah good call XD whatever is on the bottom is your `u` :3
okay thanks a lot!

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