Problem 42 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{d x}{x^{3} \sqrt{x^{2}-100}}, x>10$$

Step-by-Step Solution

Answer

#tag_title#Step 4: Evaluate the individual integrals and combine#tag_content# For the first integral: $$\int \cos^2{t} d t = \int \frac{1 + \cos{2t}}{2} d t = \frac{1}{2} \int (1 + \cos{2t}) d t = \frac{1}{2} (t + \frac{1}{2}\sin{2t}) + C_1.$$ For the second integral: $$\int \frac{\cosh^2{u}}{\cosh^2{u} \cosh{u}}d u - \int \frac{1}{\cosh^2{u} \cosh{u}} d u = \int \frac{1}{\cosh{u}} d u - \int \frac{1}{\cosh^3{u}} d u.$$ The first part can be solved directly: $$\int \frac{1}{\cosh{u}} d u = \int \frac{\cosh{u}}{\cosh^2{u}} d u = \int \frac{\cosh{u}}{\cosh^2{u} - \sinh^2{u}} d u = \int \operatorname{sech}u \,d u.$$ This integral can be solved, and its result is: $$\int \operatorname{sech}u \,d u = \arctan(\sinh u) + C_2.$$ Now, let's rewrite the result using the substitution we made at the beginning: $$\int \operatorname{sech}u \,d u = \arctan(\sinh u) + C_2 = \arctan\left(\frac{x}{10}\sqrt{1 - \frac{100}{x^2}} \right) + C_2.$$ Now, we can combine the results of the first and second integral: $$\int \frac{1}{\cosh^{3}{u}}d u = \frac{1}{2} (t + \frac{1}{2}\sin{2t}) + C_1 - \left(\arctan\left(\frac{x}{10}\sqrt{1 - \frac{100}{x^2}} \right) + C_2\right).$$ Finally, we can change the variable \(t\) back to \(u\), using \(\cosh{u} = \sec{t}\) so that \(\arccos{(\cosh{u})} = t\), to get the final answer: $$\int \frac{1}{x^{3} \sqrt{x^{2}-100}}d x = \frac{1}{2} \left(\arccos{(\cosh{u})} + \frac{1}{2}\sin{2 \arccos{(\cosh{u})}}\right) - \arctan\left(\frac{x}{10}\sqrt{1 - \frac{100}{x^2}} \right) + C_3.$$ So the final answer is: $$\int \frac{1}{x^{3} \sqrt{x^{2}-100}}d x = \frac{1}{2} \left(\arccos{(\cosh{u})} + \frac{1}{2}\sin{2 \arccos{(\cosh{u})}}\right) - \arctan\left(\frac{x}{10}\sqrt{1 - \frac{100}{x^2}} \right) + C.$$

Step 1: Determine the substitution for the integral

Let's try to find a substitution that simplifies the integral further. A common technique is to use a hyperbolic trigonometric function. Let's use the substitution: $$x = 10 \cosh{u},$$ since this substitution gives us \((x^2 - 100) = (10\sinh{u})^2\), which simplifies the square root in the denominator nicely. Now, we must find the derivative of \(x\) with respect to \(u\) which will be \(\frac{dx}{du}\).

Step 2: Calculate the derivative of the substitution and substitute

Using the substitution, we want to find \(\frac{dx}{du}\), which is the derivative of \(x\) with respect to \(u\). We differentiate \(x = 10 \cosh{u}\) with respect to \(u\) and get: $$\frac{dx}{du} = 10 \sinh{u}.$$ Now, we can rewrite \(dx\) as: $$dx = 10\sinh{u} d u.$$ Substitute the expressions for \(x\) and \(dx\) into the integral: $$\int \frac{d x}{x^{3} \sqrt{x^{2}-100}} = \int \frac{10\sinh{u}d u}{(10 \cosh{u})^{3} \sqrt{(10 \cosh{u})^{2}-100}}.$$

Step 3: Simplify the integral and solve

Now, we can simplify the integral: $$\int \frac{10\sinh{u}d u}{(10 \cosh{u})^{3} \sqrt{(10 \cosh{u})^{2}-100}} = \int \frac{10\sinh{u}d u}{(10^3 \cosh^{3}{u}) \sqrt{100\sinh^2{u}}} = \int\frac{\sinh u }{\cosh^{3}u \sinh{u}}du = \int\frac{1}{\cosh^{3}{u}}du.$$ Now the integral becomes: $$\int \frac{1}{\cosh^{3}{u}}d u.$$ To solve this integral, we will use a trigonometric identity: $$\cosh^2{u} - \sinh^2{u} = 1.$$ Now, we can rewrite the integral as: $$\int \frac{1 - \sinh^2{u}}{\cosh^2{u} \cosh{u}}d u.$$ Now we can split the integral into two integrals: $$\int \frac{1}{\cosh^{3}{u}}d u = \int \frac{1}{\cosh^{2}{u}\cosh{u}}d u - \int \frac{\sinh^2{u}}{\cosh^2{u} \cosh{u}}d u.$$ You can use secant and tangent substitution (Let \(\cosh{u} = \sec{t}\)) for the first integral: $$\int \frac{1}{\cosh^{2}u\cosh{u}}d u = \int \frac{1}{\sec^{2}{t}\sec{t}}d t = \int \cos^2{t} d t.$$ For the second integral, we can simplify it as $$\int \frac{\sinh^2{u}}{\cosh^2{u} \cosh{u}}d u = \int \frac{\cosh^2{u} - 1}{\cosh^2{u} \cosh{u}}d u = \int \frac{\cosh^2{u}}{\cosh^2{u} \cosh{u}}d u - \int \frac{1}{\cosh^2{u} \cosh{u}} d u.$$ Now it remains to evaluate these individual integrals and combine them to get the final result.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique in integration used to simplify complex expressions, particularly those involving square roots. It's especially useful when dealing with integrals containing terms like \(\sqrt{x^2 - a^2}\), \(\sqrt{x^2 + a^2}\), or \(\sqrt{a^2 - x^2}\). The method involves substituting a trigonometric function in place of the variable, which turns the expression into a trigonometric identity.

For example, to handle the integral \(\int \frac{d x}{x^{3} \sqrt{x^{2}-100}}\), one might normally consider substituting \(x = a \sec{\theta}\) or \(x = a \tan{\theta}\), based on the form of the square root. This substitution converts the square root into a simpler trigonometric form, which can be integrated more easily. This step is crucial in turning a challenging integral into a more manageable one.

Hyperbolic Functions

Hyperbolic functions, analogous to trigonometric functions, offer a remarkable way to simplify complex integrands, especially those involving expressions like \(x^2 - a^2\). These functions are defined using exponential functions, with \(\cosh{u} = \frac{e^u + e^{-u}}{2}\) and \(\sinh{u} = \frac{e^u - e^{-u}}{2}\) being two of the most common ones, similar to cosine and sine.

In our exercise, swapping trigonometric with hyperbolic functions provides a smoother path to simplify terms under square roots. When substituting \(x = 10 \cosh{u}\), the square root \((x^2 - 100)\) turns into \((10\sinh{u})^2\), which is much easier to integrate. The use of hyperbolic identities like \(\cosh^2{u} - \sinh^2{u} = 1\) further aids in reducing the integrals to simpler forms.

Integral Calculus

Integral calculus, a cornerstone of calculus alongside differential calculus, is focused on accumulation and area determination. It connects the derivative of a function with its integral and plays a crucial role in understanding the geometric and physical properties of spaces and materials.

The exercise showcases a fascinating aspect of integral calculus where, through various substitution methods, we transform a complex integral into one solvable by simpler calculus techniques. Understanding and mastering these methods opens the door to solving real-world problems involving areas, volumes, and growth rates, highlighting the multi-faceted applications of calculus.

Mathematical Substitution

Mathematical substitution is a versatile technique in calculus to simplify expressions and make integrals more tractable. By introducing a new variable, typically a function of the original variable, this method allows us to rewrite integrals in a form where the solution becomes more straightforward.

In this context, the substitution \(x = 10 \cosh{u}\) transforms the original integral into one in terms of the new variable \(u\). The integration \(\int \frac{d x}{x^{3} \sqrt{x^{2}-100}}\) then simplifies dramatically, permitting the computation of the integral in stages. By substituting and recalculating derivatives, integration becomes accessible, showcasing the technique's effectiveness in handling intricate problems in calculus.