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Chapter 7: Problem 48

Use a computer algebra system to evaluate the following indefinite integrals. Assume that a is a positive real number. $$\int \sqrt{4 x^{2}+36} d x$$

Short Answer

Expert verified
Short Answer: To solve the integral \(\int \sqrt{4 x^{2}+36} d x\), we use the substitution \(x = \frac{3}{2}\sinh(u)\), which ultimately gives us the antiderivative \(-\frac{1}{4}\sinh(2u) + \frac{3}{2}x\cosh(u) + C\), where C is a constant of integration.

Step by step solution

01

Identify the hyperbolic trigonometric identity

The given integral reminds us of the hyperbolic trigonometric identity: $$\cosh^2(t) - \sinh^2(t) = 1$$
02

Make a substitution

Rewrite the integral in terms of the hyperbolic functions by making the substitution: $$x = \frac{3}{2}\sinh(u)$$ Then, we have: $$dx = \frac{3}{2}\cosh(u)du$$ Substituting into the integral: $$\int \sqrt{4 x^{2}+36} d x = \int \sqrt{4(\frac{3}{2}\sinh(u))^2 + 36} \cdot \frac{3}{2}\cosh(u) du$$
03

Simplify the integral

Simplify the expression inside the square root and the integral becomes: $$\int \sqrt{4(\frac{3}{2}\sinh(u))^2 + 36} \cdot \frac{3}{2}\cosh(u) du = \int \sqrt{9\sinh^2(u) + 36}\cdot\frac{3}{2}\cosh(u) du$$ Now we make use of the identity \(9\sinh^2(u) + 36 = 36\cosh^2(u)\): $$\int \sqrt{9\sinh^2(u) + 36}\cdot\frac{3}{2}\cosh(u) du = \int \sqrt{36\cosh^2(u)}\cdot\frac{3}{2}\cosh(u) du$$
04

Integrate with respect to u

The integral becomes: $$\int \sqrt{36\cosh^2(u)}\cdot\frac{3}{2}\cosh(u) du = \int 6\cosh^2(u)\cdot\frac{3}{2}\cosh(u) du$$ Now integrating with respect to u, we have: $$\int 9\cosh^3(u) du = \frac{9}{4}u\cosh^2(u)-\frac{9}{4}\int u\cdot 2\sinh(u)\cosh(u) du$$
05

Perform integration by parts

Now we need to perform integration by parts: Let \(v = u\), then \(dv = du\) Let \(dw = 2\sinh(u)\cosh(u) du\), then \(w = \sinh^2(u)\) Now we have: $$-\frac{9}{4}\int u\cdot 2\sinh(u)\cosh(u) du = -\frac{9}{4}\left[u\sinh^2(u)-\int \sinh^2(u) du\right]$$ To solve the remaining integral, use the identity \(\sinh^2(u) = \frac{1}{2}\left(\cosh(2u) - 1\right)\). The integral is now: $$-\frac{9}{8}\left[u\sinh^2(u)-\frac{1}{2}\int (\cosh(2u) - 1) du\right]$$ Integrating: $$-\frac{9}{8}\left(u\sinh^2(u)-\frac{1}{4}\sinh(2u) + \frac{1}{2}u\right)$$ Now we have two parts of the antiderivative: $$\frac{9}{4}u\cosh^2(u) -\frac{9}{8}\left(u\sinh^2(u)-\frac{1}{4}\sinh(2u) + \frac{1}{2}u\right)$$
06

Substitute back for x

Rewrite our antiderivative in terms of x using the substitution \(x = \frac{3}{2}\sinh(u)\), and note that \(\cosh^2(u) = \frac{4}{9}\left(\frac{4x^2}{9} + 1\right)\) The final answer is: $$\frac{9}{4}u\cosh^2(u) -\frac{9}{8}\left(u\sinh^2(u)-\frac{1}{4}\sinh(2u) + \frac{1}{2}u\right) = -\frac{1}{4}\sinh(2u) + \frac{3}{2}x\cosh(u)$$

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