/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 25

Evaluate the following integrals. $$\int \frac{d x}{\sqrt{x^{2}-81}}, x>9$$

Short Answer

Expert verified
The integral of the function \(f(x) = \frac{1}{\sqrt{x^2-81}}\) under the condition \(x > 9\) is: $$\int \frac{dx}{\sqrt{x^2 - 81}} = \cosh^{-1}\left(\frac{x}{9}\right) + C$$

Step by step solution

01

Trigonometric Substitution

Substitute \(x\) with \(9\cosh u\): $$x = 9\cosh u \Rightarrow dx = 9\sinh u \, du$$ Now substitute these expressions into the integral: $$\int \frac{1}{\sqrt{(9\cosh u)^2 - 81}} \cdot 9\sinh u \, du$$
02

Simplifying the Integral

Simplify the integral by factoring out \(81\) from the square root expression and canceling out terms: $$\int \frac{1}{\sqrt{81(\cosh^2 u - 1)}} \cdot 9\sinh u \, du = \int \frac{1}{\sqrt{81\cdot9^2(\sinh^2 u)}} \cdot 9\sinh u \, du$$ Since \(\sqrt{81} = 9\) and \(\sqrt{9^2} = 9\): $$\int \frac{1}{9^2 \sqrt{\sinh^2 u}} \cdot 9\sinh u \, du = \int \frac{1}{9\sqrt{\sinh^2 u}} \cdot 9\sinh u \, du$$ Cancel out the \(9\) terms: $$\int \frac{\sinh u}{\sqrt{\sinh^2 u}} \, du$$
03

Integrating the new expression

The expression inside the integral is now easier to integrate: $$\int \frac{\sinh u}{\sqrt{\sinh^2 u}} \, du = \int \frac{\sinh u}{\left| \sinh u \right|} \, du$$ Given that \(x > 9\), which means \(u > 0\) because \(\cosh u > 1\) for any \(u\), the expression inside the integral becomes: $$\int \frac{\sinh u}{\sinh u} \, du = \int 1 \, du$$ Integrating with respect to \(u\): $$\int 1 \, du = u + C$$
04

Replacing the original variable

Now we must write the result using the original variable \(x\), using the substitution we made earlier. Since \(x = 9\cosh u\), we find \(u\) in terms of \(x\): $$u = \cosh^{-1} \left(\frac{x}{9}\right)$$ Putting that into our result, we get: $$u + C = \cosh^{-1}\left(\frac{x}{9}\right) + C$$ Thus, the final solution is: $$\int \frac{dx}{\sqrt{x^2 - 81}} = \cosh^{-1}\left(\frac{x}{9}\right) + C$$

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