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calculus
h(x)= integral from (1, 1/x) arctan(2t)dt part 1: let U= 1/x and du= ? > using u=1/x, we can write h(x)= integral from (1, 1/x) arctan (2t)dt as h(u)= integral from (1,u) arctan(2t)dt and h'(u)= arctan (2) Part 2: By the chain
asked by Anonymous on July 14, 2011 
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also: integral of tan^(1)y dy how is integration of parts used in that? You write: arctan(y)dy = d[y arctan(y)]  y d[arctan(y)] Here we again have used the product rule: d(fg) = f dg + g df You then use that: d[arctan(y)] =
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let the function h(x)= (integrand symbol from 2 to x^2)arctan (t) dt. Find h'(x). This question confused me because i know the derivative of an integral is the original function. I just need help with finding the derivative of
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Arrange these in order from least to greatest: arctan(sqrt3), arctan 0, arctan(1/2) So far I got the first two values, arctan(sqrt3), and that's 150 degrees. Arctan 0 would be zero degrees. I'll use just one answer for now, but
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Let f be a function defined by f(x)= arctan x/2 + arctan x. the value of f'(0) is? It's 3/2 but I am not very clear on how to obtain the answer. I changed arctan x/2 into dy/dx=(42x)/(4sqrt(4+x^2)) but that's as far as I got.
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Note that pi lim arctan(x ) =  x > +oo 2 Now evaluate / pi \ lim arctan(x )   x x > +oo \ 2 / I'm not exactly sure how to attempt it. I have tried h'opital's rule but I don't believe you can use it here. Any help
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The values of x that are solutions to the equation cos^(2)x=sin2x in the interval [0, pi] are a. arctan(1/2) only b. arctan(1/2) and pi c. arctan(1/2) and 0 d. arctan(1/2) and (pi/2) e. arctan(1/2), o, and (pi/2)
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Use the integral identity: ∫(a1) (1/(1+x^2))dx=∫(11/a) (1/(1+u^2))du for a>1 to show that: arctan(a)+arctan(1/a)=π/2
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