How to Integrate 1-cos(t)dt and Prove the Resulting Derivative

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To integrate 1-cos(t)dt, the integral can be split into two parts: the integral of 1 and the integral of -cos(t). The integral of 1 with respect to t is t, and the integral of -cos(t) is -sin(t), leading to the result g(x) = x - sin(x) + C. The Fundamental Theorem of Calculus confirms that the derivative g'(x) equals 1 - cos(x). Understanding these basic integrals and their derivatives is crucial for solving the problem effectively.
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Homework Statement



it's been a long summer and i forgot everything.
i need help integrating 1-cos(t)dt

evaluate g(x) = [\pi,x]integral (1-cos(t))dt

i know by pt2 of fund. thm of calc that g' (x) = 1-cos(x)
but i can't remember how to evaluate the integral and then differentiate to prove my answer.\
I should just know the integral of 1-cosx but I'm retarded and forgot what to do.
any help?
 
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Well you know that by the properties of integrals, you can treat it as two integrals
I=I_1+I_2=\left(\int 1dt\right)+\left(-\int\cos{t}dt\right)
Do you know these integrals? What is the derivative of -\sin{t}?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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