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Chapter 7: Problem 37

Evaluate the following integrals. $$\int \frac{\sqrt{9 x^{2}-25}}{x^{3}} d x, x>\frac{5}{3}$$

Short Answer

Expert verified
Question: Evaluate the integral $$\int \frac{\sqrt{9 x^{2}-25}}{x^{3}} d x.$$ Answer: $$\int \frac{\sqrt{9 x^{2}-25}}{x^{3}} d x = 9\left(\frac{\sqrt{9x^2 - 25}}{5x} - \frac{1}{2}\left(\frac{\sqrt{9x^2 - 25}}{5x}\right)^3\right) + C$$

Step by step solution

01

Identify the trigonometric substitution

To simplify the expression under the square root, we should find a trigonometric substitution such that the denominator becomes equal to a constant. In this case, the substitution to use is: $$x = \frac{5}{\sqrt{3}} \cosh{\theta} \quad \text{with} \quad x>\frac{5}{3}$$ which fulfills the desired condition.
02

Calculate the differential dx

Now, we need to find the expression for the differential dx in terms of d胃. This can be found by differentiating x w.r.t 胃: $$\frac{dx}{d\theta} = \frac{5}{\sqrt{3}} \sinh{\theta}.$$ So, $$dx = \frac{5}{\sqrt{3}} \sinh{\theta} d\theta.$$
03

Perform the substitution

Using the substitution and its differential, we can rewrite the integral as: $$ \int \frac{\sqrt{9 (\frac{5}{\sqrt3}\cosh{\theta})^2 - 25}}{(\frac{5}{\sqrt{3}} \cosh{\theta})^3} \cdot \frac{5}{\sqrt{3}} \sinh{\theta} d\theta = \int \frac{\sqrt{9 \cdot \frac{25}{3} \cosh^2{\theta} - 25}}{ \frac{125}{9} \cosh^3{\theta}} \cdot \frac{5}{\sqrt{3}} \sinh{\theta} d\theta. $$
04

Simplify the integrand

Now we can simplify the integral: $$ \int \frac{\sqrt{9 \cdot \frac{25}{3} \cosh^{2}{\theta} - 25}}{ \frac{125}{9} \cosh^{3}{\theta}} \cdot \frac{5}{\sqrt{3}} \sinh{\theta} d\theta = \int \frac{\frac{15}{\sqrt3 }\sqrt{\cosh^{2}{\theta} - 1}}{\frac{125}{9} \cosh^{3}{\theta}} \cdot \frac{5}{\sqrt{3}} \sinh{\theta} d\theta. $$ We can simplify further by using the identity: $$\sinh^2{\theta} = \cosh^2{\theta} - 1$$ So, the integral becomes: $$ \int \frac{\frac{15}{\sqrt3} \cdot \sinh{\theta}}{ \frac{125}{9} \cosh^{3}{\theta}} \cdot \frac{5}{\sqrt{3}} \sinh{\theta} d\theta = 9\int \frac{\sinh^2{\theta}}{\cosh^{3}{\theta}} d\theta. $$
05

Perform the integration

Now we can integrate with respect to 胃: $$ 9\int \frac{\sinh^2{\theta}}{\cosh^{3}{\theta}} d\theta = 9\int \frac{(\cosh^2{\theta} - 1)}{\cosh^{3}{\theta}} d\theta = 9\int (\frac{1}{\cosh{\theta}} - \frac{1}{\cosh^3{\theta}}) d\theta. $$ Using the properties of hyperbolic functions, we find the antiderivative: $$ 9\int (\frac{1}{\cosh{\theta}} - \frac{1}{\cosh^3{\theta}}) d\theta = 9(\tanh{\theta} - \frac{1}{2}\tanh^3{\theta}) + C. $$
06

Reverse the substitution

Now, we need to express our result in terms of x. Recall the original substitution: $$x = \frac{5}{\sqrt{3}} \cosh{\theta} \quad \implies \quad \tanh{\theta} = \frac{\sqrt{9x^2 - 25}}{5x}.$$ Thus, our final answer is: $$ 9(\tanh{\theta} - \frac{1}{2}\tanh^3{\theta}) + C = 9\left(\frac{\sqrt{9x^2 - 25}}{5x} - \frac{1}{2}\left(\frac{\sqrt{9x^2 - 25}}{5x}\right)^3\right) + C $$ So, the evaluated integral is: $$ \int \frac{\sqrt{9 x^{2}-25}}{x^{3}} d x = 9\left(\frac{\sqrt{9x^2 - 25}}{5x} - \frac{1}{2}\left(\frac{\sqrt{9x^2 - 25}}{5x}\right)^3\right) + C $$

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