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

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).
Show that

$$\int _{\sqrt{2}} ^2 \frac{1}{t^3 \sqrt{t^2 - 1}} \: dt = \frac{1}{24}\left( -6 + 3\sqrt{3} + \pi \right)$$

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

$$t=\sec \theta \Rightarrow \frac{dt}{d\theta}=\sec \theta \tan \theta \Rightarrow dt = \sec \theta \tan \theta \: d\theta$$

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

$$\int _{\sec \sqrt{2}} ^{\sec 2} \frac{\sec \theta \tan \theta}{\sec ^3 \theta \sqrt{\sec ^2 \theta - 1}} \: d\theta = \int _{\sec \sqrt{2}} ^{\sec 2} \frac{\tan \theta}{\sec ^2 \theta \sqrt{\tan ^2 \theta}} \: d\theta$$

$$\int _{\sec \sqrt{2}} ^{\sec 2} \frac{\tan \theta}{\sec ^2 \theta \sqrt{\tan ^2}} \: d\theta = \int _{\sec \sqrt{2}} ^{\sec 2} \frac{d\theta}{\sec ^2 \theta} = \int _{\sec \sqrt{2}} ^{\sec 2} \cos ^2 \theta \: d\theta$$

$$\int _{\sec \sqrt{2}} ^{\sec 2} \cos ^2 \theta \: d\theta = \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} (1+\cos 2\theta) \: d\theta = \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \: d\theta + \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \cos 2\theta \: d\theta$$

Consider the following

$$\int \cos 2x \: dx$$

$$u=2x \Rightarrow \frac{du}{dx}= 2 \Rightarrow dx = \frac{du}{2}$$

$$\int \cos 2x \: dx = \frac{1}{2} \int \cos u \: du = \frac{1}{2} \sin u + \mathrm{C} = \frac{1}{2} \sin 2x + \mathrm{C} = \sin x \cos x + \mathrm{C}$$

Then, we get

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

which is wrong.

Any help is highly appreciated.

Of course,the limits of integration are incorrect.

$$t=2\Rightarrow \theta=\mbox{arcsec} \ 2$$

$$t=\sqrt{2} \Rightarrow \theta=\mbox{arcsec} \ \sqrt{2}$$

Daniel.

Oops! Thanks for pointing it out.

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

## 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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