/*! 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 29 Find the integral. $$ \int \... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 29

Find the integral. $$ \int \operatorname{arcsec} 2 x d x, \quad x>\frac{1}{2} $$

Short Answer

Expert verified
The integral of \( \operatorname{arcsec} 2x dx \) is \( x \operatorname{arcsec} 2x + \frac{1}{2} \ln |2x + \sqrt{4x^2 - 1}| + C \)

Step by step solution

01

Recognize the integral structure

Observe the given integral \( \int \operatorname{arcsec} 2x dx \), and let \( u = 2x \). Substituting this into the integral gives us \( \frac{1}{2} \int \operatorname{arcsec} u du \)
02

Apply the formula

Now substitute the formula for this integral which is: \( \int \operatorname{arcsec} u du = u \operatorname{arcsec} u + \ln |u + \sqrt{u^2 - 1}| + C \). Substitute \( u = 2x \) back into the formula to give us: \( \frac{1}{2} [2x \operatorname{arcsec} 2x + \ln |2x + \sqrt{4x^2 - 1}|] + C \)
03

Simplify the integral

Simplify the expression to get the final result. This gives us: \( x \operatorname{arcsec} 2x + \frac{1}{2} \ln |2x + \sqrt{4x^2 - 1}| + C \)

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Most popular questions from this chapter

Graphical Analysis In Exercises \(\mathbf{6 1}\) and 62, graph \(f(x) / g(x)\) and \(f^{\prime}(x) / g^{\prime}(x)\) near \(x=0 .\) What do you notice about these ratios as \(x \rightarrow 0\) ? How does this illustrate L'H么pital's Rule? \(f(x)=\sin 3 x, \quad g(x)=\sin 4 x\)

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