Problem 35 - Calculus: One and Several Variables

Question

Use integration by parts to derive the formula. $$\int \operatorname{arcsec} x d x=x \operatorname{arcsec} x-\ln |x+\sqrt{x^{2}-1}|+C.$$

Step-by-Step Solution

Answer

To derive the integral of \(\operatorname{arcsec}(x)\) using integration by parts, we choose \(u = \operatorname{arcsec}(x)\) and \(dv = dx\). We find \(du = \dfrac{1}{|x|\sqrt{x^2 - 1}} dx\) and \(v = x\). Applying the formula \(\int u dv = uv - \int v du\), we get \(\int \operatorname{arcsec}(x) dx = x \operatorname{arcsec}(x) - \int x \dfrac{1}{|x|\sqrt{x^2 - 1}} dx\). After simplification, we obtain the final formula: \[\int \operatorname{arcsec}(x) dx = x \operatorname{arcsec}(x) - \ln \left|x + \sqrt{x^2 - 1}\right| + C\]

Step 1: Identify and choose the functions u and dv

To derive the integral of \(\operatorname{arcsec}(x)\) using integration by parts, we need to choose \(u\) and \(dv\) such that we can easily compute their derivatives and integrals. We choose the following functions: \(u = \operatorname{arcsec}(x)\) and \(dv = dx\)

Step 2: Compute the derivatives of u and the integrals of dv

Now, we calculate \(du\) and \(v\): For \(u\), we need to calculate the derivative, which is: \[du = \dfrac{1}{|x|\sqrt{x^2 - 1}} dx\] For \(dv\), we need to calculate the integral, which is: \[v = \int dx = x\]

Step 3: Apply the integration by parts formula

Using the functions we have chosen for \(u\), \(dv\), \(du\), and \(v\), we plug them into the integration by parts formula: \[\int \operatorname{arcsec}(x) dx = \int u dv = uv - \int v du\] Substitute the values, we get: \[\int \operatorname{arcsec}(x) dx = x \operatorname{arcsec}(x) - \int x \dfrac{1}{|x|\sqrt{x^2 - 1}} dx\]

Step 4: Simplify the integral and perform integration

Now, we will simplify and solve the remaining integral: \[\int \operatorname{arcsec}(x) dx = x \operatorname{arcsec}(x) - \int \dfrac{x}{|x|\sqrt{x^2 - 1}} dx\] We can simplify the integrand as follows: \(\dfrac{x}{|x|\sqrt{x^2 - 1}} = \dfrac{1}{\sqrt{x^2 - 1}}\), since \(\dfrac{x}{|x|}\) is either 1 or -1 depending on the sign of x. Now, the integral becomes: \[\int \operatorname{arcsec}(x) dx = x \operatorname{arcsec}(x) - \int \dfrac{1}{\sqrt{x^2 - 1}} dx\] We should recognize this integral as the standard formula for the integral of an inverse hyperbolic function: \(\int \dfrac{1}{\sqrt{x^2 - 1}} dx = \ln \left|x + \sqrt{x^2 - 1}\right| + C\)

Step 5: Write down the final formula

Now, we combine these results to obtain the final formula: \[\int \operatorname{arcsec}(x) dx = x \operatorname{arcsec}(x) - \ln \left|x + \sqrt{x^2 - 1}\right| + C\] And that's the derived integral formula for the arcsec function using integration by parts.

Key Concepts

Arcsecant Integration

Integration is a core concept in calculus, which comes in handy for many mathematical and real-world applications. In particular, integrating the arcsecant function, denoted \(\operatorname{arcsec} x\), is an intriguing exercise that involves the technique of integration by parts.

Integration by parts is a method derived from the product rule in differentiation and serves as a powerful tool for solving integrals that are products of two functions. For arcsecant integration, one typically sets the \(\operatorname{arcsec} x\) function as \(u\) and the differential \(dx\) as \(dv\). The process involves finding the derivative of \(u\) and the integral of \(dv\). Subsequent steps require strategic algebraic manipulations and recognizing familiar integration patterns, such as the integral of an inverse hyperbolic function.

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverses of the trigonometric functions and are key players in calculus. They allow us to find an angle when given its trigonometric ratio.

The main inverse trigonometric functions are arcsine \(\operatorname{arcsin}\), arccosine \(\operatorname{arccos}\), and arcsecant \(\operatorname{arcsec}\). Each of these functions is defined on specific domain intervals to keep them as functions, meaning that each input yields only one output. They often surface in integrals, especially when dealing with geometrical contexts or when inverting the process of differentiation of trigonometric functions. Understanding their properties, like their derivatives, is essential to perform integration by parts effectively.

Calculus

Calculus is a vast field of mathematics that studies continuous change and it's subdivided into differential and integral calculus.

Differential Calculus

It focuses on the rate at which quantities change. The fundamental process in this subdivision is differentiation, a process that finds how a function changes infinitesimally.

Integral Calculus

Conversely, integral calculus looks at the accumulation of quantities, such as areas under curves. Integration is the central process here, and it can be viewed as an inverse operation to differentiation. Examples in various scientific and engineering fields where calculus is applied range from computing orbits in celestial mechanics to optimizing functions in economics.

Integral Calculus

Integral calculus is an integral part of mathematics, pun intended, and is concerned with the accumulation of quantities, such as area under a curve, volume under a surface, and other physical quantities like mass and charge.

Techniques such as substitution, integration by parts, and partial fractions expand the arsenal of functions that can be integrated. Integration by parts stands out as it transforms the integral of a product of functions into a (hopefully) simpler problem to tackle. Mastering this technique opens the door to solving integrals involving products of algebraic, exponential, logarithmic, and trigonometric functions, including the challenges presented by the arcsecant function.