just finished chapter 5 (intro to integration) from stewart, could you guys quiz me on it? i think integration of trig functions is what I'm worst at because I cannot nail the substitution of the range parameters when substituting terms in the expression
[eqn] \int_{-\infty}^\infty \frac{\cos x}{x^2+1} \text{d}x =? [/eqn]
Theres no point in learning more than you need.Imagine if your job consisted of using MS Word and you decided to figure out what every button and keyboard shortcut does.
>>16999333>learning is stupid if it doesn't directly benefit me!
>>16999333it's pretty straightforward really, and advanced word users know all its capabilities
>>16999329This is a nice one. I'll start with ~1.4 It damped pretty fast, I would have sketched a few more oscillations. Do we get a cheat sheet with multiplication rule and identities and stuff?
>>16999406Are you a retard? Solving integrals with trig identities is not 'type it into graphing software' faggot go open a textbook
>>16999414Show your work
>>16999434step 1. looked at itstep 2. thought about it for 5 secondsstep 3. wrote down the answer, it's [math] \pi/e [/math]
>>16999438I'll guess wolfram alpha? 1.16 is pretty close to 1.4, I could have estimated better
>>16999445No, just notice that the value if the integral doesn't change if you replace the cos(x) by exp(ix) (the sine component is odd). Then just extend the integration contour to a semicircle in the upper half plane and apply the residue theorem at z=i
>>16999449Interesting method, I'm not totally convinced that you haven't just recited a known solution, and it's probably beyond chapter 5 of a calculus 1 book
>>16999306[math]\int_0^{a/3}\frac{dx}{\sqrt{a^2-x^2}}[/math]
>>16999449It's not really obvious to me how this is more convenient...
>>16999449But, I did find what I think you're reciting :)
