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How do you integrate one over sqrt of x squared minus nine between limits five and ten?
5∫10[1/√(x2 - 9)] dx = 5∫10[dx/√(x2 - 32)]
Let x = 3 sec θ, where 0 < θ < π/2 or π < θ < 3π/2. Then, dx = 3 sec θ tan θ dθ and
√(x2 - 9) = √[(3 sec θ)2 - 9] = √[9(sec2 θ - 1)] = √(9 tan2 θ) = 3 tan θ
5∫10[dx/√(x2 - 32)]
= 5∫10[(√3 sec θ tan θ)/(3 tan θ)] dθ
= 5∫10 sec θ dθ
= ln |sec θ + tan θ|5|10 = ln sec 10 + tan 10 - sec 5 - tan 5 = -2.298
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