Integrate \frac{1}{t^3 \sqrt{t^2 - 1}} from \sqrt{2} to 2

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In summary, the conversation discusses the use of trigonometric substitution to solve the integral \int _{\sqrt{2}} ^2 \frac{1}{t^3 \sqrt{t^2 - 1}} \: dt and the correct limits of integration being \mbox{arcsec } \sqrt{2} and \mbox{arcsec } 2. The final solution is \frac{1}{24}\left( -6 + 3\sqrt{3} + \pi \right).
  • #1
DivGradCurl
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[tex]\int _{\sqrt{2}} ^2 \frac{1}{t^3 \sqrt{t^2 - 1}} \: dt = \frac{1}{24}\left( -6 + 3\sqrt{3} + \pi \right)[/tex]

I've tried to use the trigonometric substitution [tex]t=\sec \theta[/tex], but without success thus far.

[tex]t=\sec \theta \Rightarrow \frac{dt}{d\theta}=\sec \theta \tan \theta \Rightarrow dt = \sec \theta \tan \theta \: d\theta[/tex]

[tex]\int _{\sqrt{2}} ^{2} \frac{1}{t^3 \sqrt{t^2 - 1}} \: dt=\int _{\sec \sqrt{2}} ^{\sec 2} \frac{\sec \theta \tan \theta}{\sec ^3 \theta \sqrt{\sec ^2 \theta - 1}} \: d\theta[/tex]

[tex]\int _{\sec \sqrt{2}} ^{\sec 2} \frac{\sec \theta \tan \theta}{\sec ^3 \theta \sqrt{\sec ^2 \theta - 1}} \: d\theta = \int _{\sec \sqrt{2}} ^{\sec 2} \frac{\tan \theta}{\sec ^2 \theta \sqrt{\tan ^2 \theta}} \: d\theta[/tex]

[tex]\int _{\sec \sqrt{2}} ^{\sec 2} \frac{\tan \theta}{\sec ^2 \theta \sqrt{\tan ^2}} \: d\theta = \int _{\sec \sqrt{2}} ^{\sec 2} \frac{d\theta}{\sec ^2 \theta} = \int _{\sec \sqrt{2}} ^{\sec 2} \cos ^2 \theta \: d\theta[/tex]

[tex]\int _{\sec \sqrt{2}} ^{\sec 2} \cos ^2 \theta \: d\theta = \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} (1+\cos 2\theta) \: d\theta = \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \: d\theta + \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \cos 2\theta \: d\theta[/tex]

Consider the following

[tex]\int \cos 2x \: dx[/tex]

[tex]u=2x \Rightarrow \frac{du}{dx}= 2 \Rightarrow dx = \frac{du}{2}[/tex]

[tex]\int \cos 2x \: dx = \frac{1}{2} \int \cos u \: du = \frac{1}{2} \sin u + \mathrm{C} = \frac{1}{2} \sin 2x + \mathrm{C} = \sin x \cos x + \mathrm{C}[/tex]

Then, we get

[tex]\frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \: d\theta + \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \cos 2\theta \: d\theta = \left. \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta \right] _{\sec \sqrt{2}} ^{\sec 2}[/tex]

which is wrong.

Any help is highly appreciated.
 
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  • #2
Of course,the limits of integration are incorrect.

[tex] t=2\Rightarrow \theta=\mbox{arcsec} \ 2 [/tex]

[tex] t=\sqrt{2} \Rightarrow \theta=\mbox{arcsec} \ \sqrt{2} [/tex]

Daniel.
 
  • #3
Oops! Thanks for pointing it out.

[tex]\left. \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta \right] _{\mbox{arcsec } \sqrt{2}} ^{\mbox{arcsec } 2} = \frac{1}{24}\left( -6 + 3\sqrt{3} + \pi \right)[/tex]
 

Related to Integrate \frac{1}{t^3 \sqrt{t^2 - 1}} from \sqrt{2} to 2

1. What does the symbol "t" represent in this integral?

The symbol "t" represents the variable of integration, which is a placeholder for any value in the given range of integration in the integral expression.

2. Why is the range of integration from √2 to 2?

The range of integration is from √2 to 2 because these are the values of "t" that make the expression inside the integral well-defined. If the expression is undefined at any point within the range, the integral cannot be evaluated.

3. How do you solve this type of integral?

To solve this type of integral, you can use the substitution method. Let u = t^2 - 1, which results in du = 2t dt. Substituting these values into the integral expression, it becomes ∫1/(t^3√u) du. You can then use the power rule and the inverse power rule to evaluate the integral.

4. What is the final answer for this integral?

The final answer for this integral is 1/2[√(2^2 - 1) - √(√2^2 - 1)] = 1/2[√3 - √1] = (1/2)√2.

5. Can this integral be solved without using substitution?

Yes, this integral can also be solved without using substitution. You can use the trigonometric substitution method, where u = √(t^2 - 1) and t = sec(θ). This results in the integral expression becoming ∫1/(u^2 + 1) du, which can be solved using the inverse tangent function.

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