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GraceLee
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∫x3
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(4x2 + 9) 3/2
According to my book this is a trig substitution integral. The normal procedure is to substitute atanθ for x when one has a square root w an argument of the form x^2 + a^2. Because the argument of the square root is 4x2 + 9, as opposed to simply x2 +9, the book suggests substituting 2x for x, hence instead of substituting 3tanθ for x- you sub 3/2tanθ. That much makes sense to me. The part that is driving me insane is that the book goes on to complete the trig substitution by substituting this x value (3/2 tanθ) for x in the numerator (ok), substituting 3/2sec2θdθ for dx (ok)... however in the denominator, they write √(4x^2 +9) = √9tan^2θ +9 = 3secθ. That's not substituting the "agreed upon" value of x (i.e., 3/2tanθ) it's just substituting 3tanθ! This is my first post and I'm no math genius so I'm certain I'm just not comprehending something but I've been stuck on this for a while and it's driving me nuts! Any help would greatly appreciated- I hope this post is fairly intelligible.
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(4x2 + 9) 3/2
According to my book this is a trig substitution integral. The normal procedure is to substitute atanθ for x when one has a square root w an argument of the form x^2 + a^2. Because the argument of the square root is 4x2 + 9, as opposed to simply x2 +9, the book suggests substituting 2x for x, hence instead of substituting 3tanθ for x- you sub 3/2tanθ. That much makes sense to me. The part that is driving me insane is that the book goes on to complete the trig substitution by substituting this x value (3/2 tanθ) for x in the numerator (ok), substituting 3/2sec2θdθ for dx (ok)... however in the denominator, they write √(4x^2 +9) = √9tan^2θ +9 = 3secθ. That's not substituting the "agreed upon" value of x (i.e., 3/2tanθ) it's just substituting 3tanθ! This is my first post and I'm no math genius so I'm certain I'm just not comprehending something but I've been stuck on this for a while and it's driving me nuts! Any help would greatly appreciated- I hope this post is fairly intelligible.