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Chapter 7: Problem 18

Use a table of integrals to determine the following indefinite integrals. $$\int \sqrt{4 x^{2}-9} d x, x>\frac{3}{2}$$

Short Answer

Expert verified
Answer: The resulting integral is \(\frac{x\sqrt{4x^2-9} - 3\cosh^{-1}\left(\frac{2x}{3}\right)}{2} + C\) where x > 3/2 and C is the constant of integration.

Step by step solution

01

Recognize the given integral form

Begin by recognizing that given integral closely resembles the hyperbolic sine function. The integral is \(\int \sqrt{4x^2-9} dx\).
02

Perform a substitution

To bring the integral to a standard form found in a table of integrals, perform the substitution \(x = \frac{3}{2}\cosh(t)\), where \(\cosh(t)\) is the hyperbolic cosine function, and compute the differential \(dx\): $$ d x = \frac{3}{2} \sinh(t) dt. $$
03

Rewrite the integral using substitution

Rewrite the integral using the substitution from Step 2, and simplify the expression: $$ \int \sqrt{4\left(\frac{3}{2}\cosh(t)\right)^2 - 9} \left(\frac{3}{2}\sinh(t)\right) \, dt. $$
04

Simplify and integrate

The expression simplifies to: $$ \int 3\sinh^2(t) \, dt. $$ Now, use the identity \(\sinh^2(t) = \frac{1}{2}(\cosh(2t) - 1)\), and the integral becomes: $$ \int \frac{3}{2} (\cosh(2t) - 1) \, dt. $$ Integrate the expression: $$ \frac{3}{4}(\sinh(2t) - 2t) + C. $$
05

Revert back to the original variable

Now that we have the integrated expression in terms of the substitution variable, revert back to the original variable \(x\). Recall that we used the substitution \(x = \frac{3}{2} \cosh(t)\). We can solve for \(t\) using the inverse hyperbolic cosine function, \(\cosh^{-1}(u)\): $$ t = \cosh^{-1}\left(\frac{2x}{3}\right). $$ The derivative of the hyperbolic sine function is the hyperbolic cosine function. Therefore, we have \(\sinh(2t) = 2\sinh(t)\cosh(t)\). Use this relationship and the inverse hyperbolic cosine to rewrite the integrated expression in terms of \(x\): $$ \frac{3}{4}\left(2\sqrt{\left(\frac{2x}{3}\right)^2 - 1}\cdot\frac{2x}{3} - 2\cosh^{-1}\left(\frac{2x}{3}\right)\right) + C. $$
06

Final answer

Write the final answer: $$ \int \sqrt{4x^2-9} dx = \frac{x\sqrt{4x^2-9} - 3\cosh^{-1}\left(\frac{2x}{3}\right)}{2} + C, $$ where x > 3/2 and C is the constant of integration.

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Most popular questions from this chapter

\(A\) powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$ F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t $$ where we assume that s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated: $$ F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1} $$ Verify the following Laplace transforms, where a is a real number. $$f(t)=e^{a t} \longrightarrow F(s)=\frac{1}{s-a}$$

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{2+\cos x}$$

Graph the function \(f(x)=\frac{1}{x \sqrt{x^{2}-36}}\) on its domain. Then find the area of the region \(R_{1}\) bounded by the curve and the \(x\) -axis on \([-12,-12 / \sqrt{3}]\) and the area of the region \(R_{2}\) bounded by the curve and the \(x\) -axis on \([12 / \sqrt{3}, 12] .\) Be sure your results are consistent with the graph.

Circumference of a circle Use calculus to find the circumference of a circle with radius \(a.\)

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1+\sin x}$$
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