Problem 29 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Recognize the integral structure

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

Step 2: 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: \( \frac{1}{2} [2x \operatorname{arcsec} 2x + \ln |2x + \sqrt{4x^2 - 1}|] + C \)

Step 3: Simplify the integral

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