<p>sq root(2 + 2cosO)</p>
<p>O=theta</p>
<p>We just learned integration a couple days ago (we learned about antiderivatives and integration by parts…but not substitution yet), but I think what you would do is:</p>
<p>sq root (2+2cosO) = (2+2cosO)^(1/2)</p>
<p>Basically, you just have to rewrite the integral in terms of u (in this case, u = 2+2cosO)
So you’d now have the integral u^(1/2)du
You can then solve that with the rule that if f(x) = x^n then F(x) = (x^(n+1))/(n+1)
So you’d have ((u)^(3/2))/(3/2)</p>
<p>And your final answer would be: (2/3)(2+cosO)^(3/2) + C = F(x)</p>
<p>You might want to have somebody verify that I did it the right way. We haven’t learned how to in class, but I just looked it up on this great site I found for Calc notes–><a href=“http://tutorial.math.lamar.edu/[/url]”>http://tutorial.math.lamar.edu/</a></p>
<p>zpmqxonw, that’s not correct. It’s not quite that simple. You’re forgetting that du does not equal dx. If you use the substitution u=2+2cosO, you get the integral of u^(1/2) / du/dx = u^(1/2) / -2Sin[O]. </p>
<p>I’m not sure how you would do this, but Mathematica gives the answer as 2 (2 + 2 Cos[θ])^(1/2) * Tan[θ/2]</p>
<p>I didn’t get an answer like Mathematica’s, but I did get an answer. Ok, my conventions: S = integral, x = theta. So, </p>
<p>S(2+2cos x)^(1/2) dx
S(2(1+ cos x))^(1/2) dx
2^(1/2)S(1+cos x)^(1/2) dx
2^(1/2)S(1+cos x)^(1/2) * (1-cos x)^(1/2)/(1-cos x)^(1/2) dx
2^(1/2)S( (1-cos^2 x)^(1/2)/(1-cos x)^(1/2) ) dx
2^(1/2)S( (sin^2 x)^(1/2)) / (1-cos x)^(1/2) dx
2^(1/2)S( sin x / (1-cos x)^(1/2) ) dx
let u = 1-cos x, then du = sin x dx
2^(1/2)S(du/u^(1/2) )
2^(1/2)[2u^(1/2)]
2^(3/2)*sqrt(1-cos x) + C</p>
<p>okay, to verify who has the correct integral i will tell you the entire problem, you Needed to integrate sq root(2 + 2cosO) from 0 to 2pi. the answer is 8 according to the back of the book. please check if your answer is 8. also someone told me you might have to use a double or half angle identity, but i dont know for sure.</p>
<p>do be able to integrate, you have to manipulate sqrt[2+2cosx].
Multiply sqrt[2+2cosx] by sqrt[2-2cosx]/sqrt[2-2cosx]. since sqrt[2-2cosx] is its conjugate, the product is sqrt[4-4(cosx)^2]/sqrt[2-2cosx].
Note that 1-(cosx)^2=(sinx)^2, thus by substitution, we get sqrt[4(sinx)^2]/sqrt[2-2cos]. This becomes 2abs[sinx]/sqrt[2-2cosx].
To get rid of the absolute value, we can take the integral of 2sinx/sqrt[2-2cosx] from 0 to pi and then multiply the answer by 2 since sinx from pi to 2pi is simply the negative of 0 to pi.
integral{2sinx/sqrt[2-2cosx]} yields 2sqrt[2-2cosx], which when evaluated from 0 to pi and multiplied by 2 yields 8.
Q.E.D.</p>
<p>arrggg!!! you guys are sooo confusing me. Here’s the answer:
sqrt (8-8cosx) + c . That’s it. And by the way, it’s the exact answer as Grim-Reaper’s
.-…/—/…-/.//-.-./.-/.-…/-.-.</p>
<p>if you asked my geometry teacher from a few years back, you would get the answer “very carefully” and she would walk away. oh how i hated her…she didnt even know math</p>
<p>you need to put the absolute value on the sinx before you integrate. If you evaluate sqrt (8-8cosx) + c from 0 to 2pi, you’ll get 0. but with the absolute value, you’ll get the correct answer, 8. i got the same answer as van1011: 2sqrt(2-2cosx) (see above to post to see how), but to consider the absolute value, you take the integral from 0 to pi and multiply the answer by 2, giving 8. (since sinx from pi to 2pi is the same from 0 to pi except negative)</p>
<p>hmmm, that’s understandable, since it’s a calculus problem, you know. >_< Anyway, once you learn about trigonometry, geometry is a loser, so probably she meant “go ask your calculus teacher”
–/./.-/-.</p>