/*! 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 24 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 24

Evaluate the following integrals. $$\int \frac{d x}{\sqrt{16+4 x^{2}}}$$

Short Answer

Expert verified
Question: Evaluate the integral $$\int \frac{d x}{\sqrt{16+4 x^{2}}}$$ Answer: $$\int \frac{d x}{\sqrt{16+4 x^{2}}} = \frac{1}{2} \ln\left|\sqrt{1 + \frac{x^2}{4}} + \frac{x}{2}\right| + C$$

Step by step solution

01

Trigonometric Substitution

In the given integral, we can recognize the quadratic polynomial \(16 + 4x^2\). We can rewrite it as \(16\left(1 + \frac{x^2}{4}\right)\). Since \(1 + \tan^2(u) = \sec^2(u)\), we can perform the substitution: \(x = 2\tan(u)\). Also, \(dx = 2\sec^2(u) du\). Rewrite the integral with this substitution: $$\int \frac{d x}{\sqrt{16 + 4 x^{2}}} = \int \frac{2\sec^2(u) du}{\sqrt{16(1 + \tan^2(u))}}$$
02

Simplification

Using our trigonometric identity, we can simplify the expression inside the square root: $$\int \frac{2\sec^2(u) du}{\sqrt{16(1 + \tan^2(u))}} = \int \frac{2\sec^2(u) du}{\sqrt{16 \sec^2(u)}} = \int \frac{2\sec^2(u) du}{4\sec(u)} = \frac{1}{2}\int \sec(u) du$$
03

Integration

Now we can integrate the resulting expression: $$\frac{1}{2}\int \sec(u) du = \frac{1}{2}(\ln|\sec(u) + \tan(u)|) + C = \frac{1}{2}\ln|\sec(u) + \tan(u)| + C$$
04

Substitution back to x

Finally, we'll substitute back our original variable \(x\). Recall that \(x = 2\tan(u)\), so \(\tan(u) = \frac{x}{2}\). Since \(\sec^2(u) = 1 + \tan^2(u)\), we can find \(\sec(u)\): $$\sec(u) = \sqrt{1 + \tan^2(u)} = \sqrt{1 + \frac{x^2}{4}}$$ Now we can substitute these back into our integral result: $$\frac{1}{2} \ln|\sec(u) + \tan(u)| + C = \frac{1}{2} \ln\left|\sqrt{1 + \frac{x^2}{4}} + \frac{x}{2}\right| + C$$ Therefore, the final solution is: $$\int \frac{d x}{\sqrt{16+4 x^{2}}} = \frac{1}{2} \ln\left|\sqrt{1 + \frac{x^2}{4}} + \frac{x}{2}\right| + C$$

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