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Chapter 1: Problem 6

Evaluate the following integrals: $$ \int \frac{\sqrt{\mathrm{y}^{2}-49}}{\mathrm{y}} d \mathrm{y} \mathrm{y}<-7 $$

Short Answer

Expert verified
Question: Evaluate the integral $$\int \frac{\sqrt{y^2-49}}{y} dy \ \ for \ \ y<-7.$$ Solution: The antiderivative of the given integral is: $$\int \frac{\sqrt{y^2-49}}{y} dy = 7\ln\left(\frac{-y}{7}\right) + C$$

Step by step solution

01

Choose a trigonometric substitution

Since the integrand contains a term of the form \(\sqrt{y^2 - 49}\), we can make the following trigonometric substitution to simplify it: $$ y = -7\cosh(u) \Rightarrow dy = 7\sinh(u) du $$ Notice we used a minus sign, as \(y<-7\).
02

Replace y and dy in the integral

Now we can substitute the expressions for \(y\) and \(dy\) we found in the previous step: $$ \int \frac{\sqrt{(-7\cosh(u))^2 - 49}}{-7\cosh(u)} (7\sinh(u) du) $$
03

Simplify the integrand

After making the substitution, we can simplify the integrand: $$ \int \frac{\sqrt{49\cosh^2(u) - 49}}{-7\cosh(u)} (7\sinh(u) du) = \int \frac{\sqrt{49(\cosh^2(u) - 1)}}{-7\cosh(u)} (7\sinh(u) du) $$ We know that \(\cosh^2(u) - 1 = \sinh^2(u)\). So we have: $$ \int \frac{\sqrt{49\sinh^2(u)}}{-7\cosh(u)} (7\sinh(u) du) = -7\int \frac{7\sinh(u)}{\cosh(u)} du $$
04

Evaluate the integral

Now, the integral becomes quite simple: $$ -7\int \frac{7\sinh(u)}{\cosh(u)} du = -7\int \tanh(u) du $$ The antiderivative of \(\tanh(u)\) is \(-\ln(\cosh(u)) + C\), so we have: $$ -7\int \tanh(u) du = -7(-\ln(\cosh(u)) + C) $$
05

Replace u with y

To get the result in terms of \(y\), we need to substitute back the original variable by using the expression \(y=-7\cosh(u)\) from step 1. To do that, notice that: $$ \cosh(u) = \frac{-y}{7} $$ So we get: $$ -7(-\ln(\cosh(u)) + C) = 7\ln\left(\frac{-y}{7}\right) + C $$ Finally, the antiderivative of the given integral is: $$ \int \frac{\sqrt{y^2-49}}{y} dy = 7\ln\left(\frac{-y}{7}\right) + C $$

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