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| Math Central | Quandaries & Queries |
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Question from Meghan, a student: Hi again, I have a question from the trigonometric substitution of my calculus course. integral of dx / (4x^2 - 25)^3/2 From what I've learned, it looks like a sqrt(u^2 - a^2) case so we should substitute u=asec(theta). However, I'm not sure how to go about solving this with the ^3/2. Thanks so much! |
Hi Meghan,
Write 4x2 - 25 as 4(x2 - 25/4) and then [ 4x2 - 25]3/2 = 43/2 [ x2 - 25/4]3/2 = 8 [ x2 - 25/4]3/2 = 8 [ x2 - (5/2)2]3/2. Now I can see that the substitution is x = 5/2 sec(θ). With this substitution the denominator becomes
[ 4x2 - 25]3/2 = 8 [ (5/2)2 sec2(θ)- (5/2)2]3/2
= 8 × [(5/2)2]3/2 [sec2(θ) - 1]3/2 = 8 × (5/2)3 [tan2(θ)]3/2
= 8 × 125/8 tan3(θ) = 125 tan3(θ)
The substitution also yields dx = 5/2 sec(θ) tan(θ) dθ.
Perform the substitution, convert the secant and tangent functions to sines and cosines and see if you can evaluate the resulting integral.
Good luck in this,
Harley
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