Question

Evaluate the integrals in Exercises 39–56.56. ∫sec(x)dx/√(ln(sec(x)+tan(x)))

Accepted Answer

Start by examining the integral: $$\int \frac{\sec(x)}{\sqrt{\ln(\sec(x) + 	an(x))}} \, dx. $$Notice the expression inside the logarithm, $$\sec(x) + 	an(x)$$, which suggests a substitution involving this term. Let’s set $$u = \ln(\sec(x) + 	an(x)). $$This substitution is motivated because the denominator contains $$\sqrt{u}$$, and the derivative of $$u$$ will likely simplify the integral. Differentiate $$u$$ with respect to $$x$$:

$$\frac{du}{dx} = \frac{1}{\sec(x) + 	an(x)} \cdot \frac{d}{dx}(\sec(x) + 	an(x)).

$$Recall that $$\frac{d}{dx} \sec(x) = \sec(x) 	an(x)$$ and $$\frac{d}{dx} 	an(x) = \sec^2$(x). $$Calculate $$\frac{d}{dx}(\sec(x) + 	an(x)) = \sec(x) 	an(x) + \sec^2$(x) = \sec(x)(	an(x) + \sec(x)).

$$Substitute back into $$\frac{du}{dx}$$:

$$\frac{du}{dx} = \frac{\sec(x)(	an(x) + \sec(x))}{\sec(x) + 	an(x)}. $$Notice that $$	an(x) + \sec(x)$$ and $$\sec(x) + 	an(x)$$ are the same, so they cancel out, leaving:

$$\frac{du}{dx} = \sec(x).

$$Therefore, $$du = \sec(x) \, dx$$, which matches the numerator of the integral. This means the integral can be rewritten in terms of $$u as$$:

$$\int \frac{1}{\sqrt{u}} \, du.

$$From here, you can proceed to integrate with respect to $$u.$$

Evaluate the integrals in Exercises 39–56.
56. ∫sec(x)dx/√(ln(sec(x)+tan(x)))

Verified video answer for a similar problem:

Key Concepts

Integration Techniques involving Substitution

Derivatives and Identities of Secant and Tangent Functions

Logarithmic Functions and their Properties