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OpenStudy (anonymous):

Integrate cos(2-t^2)

OpenStudy (anonymous):

I don't think you can integrate that...

OpenStudy (anonymous):

I didn't either but apparently its integratable....

OpenStudy (anonymous):

@Euler271 what do you think?

OpenStudy (anonymous):

it isn't integratable using standard calculus methods since it is not an elementary function. who told you it was? here is what the answer is: http://www.wolframalpha.com/input/?i=integrate+cos%282-+t%5E2%29

OpenStudy (anonymous):

well honestly I assumed it was doable because it was on one of my tests in the past and I had no idea where to start

OpenStudy (anonymous):

it must have been slightly different, like t*cos(2 - t^2), or not calculus

OpenStudy (anonymous):

no it was definitely cos(2-t^2)

OpenStudy (anonymous):

ok well thanks anyway

OpenStudy (anonymous):

if its possible, could you help me on this question too? its- If dx/dt=(2t+1)/2x and x(1)=3 find x(0). I know how to separate them but im not sure what the next step is

OpenStudy (anonymous):

\[x dx = \frac{ (2t+1) }{ 2 } dt\]integrate both sides\[\frac{ x^2 }{ 2 } = \frac{ t^2 + t }{ 2 } + C_1\]isolate x\[x(t) = \pm \sqrt{t^2 + t + C_2}\]let x(1) = 3 \[3 = \pm \sqrt{1 + 1 + C_2}\]solve for C. since x(1) = +3, we an deduce that + is the proper sign (not minus).\[9 = 2 + C_2\]it's 7. plug it back in:\[x(t) = \sqrt{t^2 + t + 7}\] to find x(0), plug t = 0\[x(0) = + \sqrt{7}\]

OpenStudy (anonymous):

ohh ok, I see it now. I should have added c when I square rooted

OpenStudy (anonymous):

yup ^_^

OpenStudy (anonymous):

thank you sooo much!

OpenStudy (anonymous):

glad i could help C:

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