/*! 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 3 Find the integral. $$ \int \... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 3

Find the integral. $$ \int \frac{1}{x \sqrt{4 x^{2}-1}} d x $$

Short Answer

Expert verified
The integral of \(\int \frac{1}{x \sqrt{4 x^{2}-1}} d x\) is \(C + sec^{-1}(2x)\)

Step by step solution

01

Identify an appropriate substitution

We can rewrite \(4x^2 - 1\) as \((2x)^2 - 1^2\), which resembles the derivative of arc secant (\(sec^{-1}\)). Hence, we perform a substitution with \(2x = sec(u)\) or \(x = \frac{sec(u)}{2}\). This yields \(dx = \frac{sec(u)tan(u)}{2} du\). We substitute these into the original integral.
02

Substitute the values into the integral

Substituting into the integral, we get:\[\int \frac{1}{\frac{sec(u)}{2}\sqrt{4(\frac{sec(u)}{2})^{2}-1}} \cdot \frac{sec(u)tan(u)}{2} du\]This simplifies to:\[\int \frac{sec(u)tan(u)}{sec(u)\sqrt{sec^2(u)-1}} du\]As tangent is the ratio of sine to cosine, and secant is the reciprocal of cosine, we can see that the integral simplifies even further to:\[\int \frac{sec(u)tan(u)}{sec(u)\sqrt{sec^2(u)-1}} du = \int \frac{tan(u)}{\sqrt{sec^2(u)-1}} du\]This is equal to: \(\int du\).
03

Evaluate the Integral

The integral of \(du\) is simply \(u + C\), where \(C\) is the constant of integration.
04

Substitute the original value back

Substituting the original value back into the equation, we replace \(u\) with \(sec^{-1}(2x)\) to get the final answer:C + \(sec^{-1}(2x)\).

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Most popular questions from this chapter

Find all the continuous positive functions \(f(x),\) for \(0 \leq x \leq\) such that \(\int_{0}^{1} f(x) d x=1, \int_{0}^{1} f(x) x d x=\alpha,\) and \(\int_{0}^{1} f(x) x^{2} d x=\alpha^{2}\) where \(\alpha\) is a real number

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Verify the differentiation formula. \(\frac{d}{d x}\left[\cosh ^{-1} x\right]=\frac{1}{\sqrt{x^{2}-1}}\)

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