Question

Recall that the substitution \(x=a \sec \theta\) implies that \(x \geq a\) (in which case \(0 \leq \theta<\pi / 2\) and \(\tan \theta \geq 0\) ) or \(x \leq-a\) (in which case \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\) ). Evaluate for \(\int \frac{\sqrt{x^{2}-1}}{x^{3}} d x,\) for \(x>1\) and for \(x<-1\)

Short answer

#Answer# For x > 1: ∫(sqrt(x²-1))/(x³) dx = ln|x| + C₁ For x < -1: ∫(sqrt(x²-1))/(x³) dx = ln|-x| + C₂

Step-by-step solution

Perform the substitution and find dx

Use the substitution x = sec(θ), where 0 ≤ θ < (π/2) (because x > 1), and a = 1. To find the dx in terms of dθ, differentiate both sides of the substitution with respect to θ: dx/dθ = d(sec(θ))/dθ = sec(θ)tan(θ)dθ Therefore, dx = sec(θ)tan(θ)dθ

Substitute x and dx in the integral and simplify

Substitute x=sec(θ) and dx=sec(θ)tan(θ)dθ into the integral: ∫(sqrt(sec²(θ)-1))/(sec³(θ)) * sec(θ)tan(θ)dθ Simplify the integrand: ∫tan(θ)dθ

Evaluate the integral

Evaluate the integral of tan(θ)dθ: ∫tan(θ)dθ = ln|sec(θ)| + C₁ where C₁ is the constant of integration.

Transform the result back into x using the inverse substitution

The inverse of the substitution x=sec(θ) is θ=sec^(-1)(x). Replace θ with sec^(-1)(x) in the result: ln|sec(sec^(-1)(x))| + C₁ = ln|x| + C₁ The integral for x > 1 is: ∫(sqrt(x²-1))/(x³) dx = ln|x| + C₁ For x < -1 (Case 2):

Perform the substitution and find dx

Use the substitution x = -sec(θ), where (π/2) < θ ≤ π (because x < -1), and a = 1. To find the dx in terms of dθ, differentiate both sides of the substitution with respect to θ: dx/dθ = d(-sec(θ))/dθ = -sec(θ)tan(θ)dθ Therefore, dx = -sec(θ)tan(θ)dθ

Substitute x and dx in the integral and simplify

Substitute x=-sec(θ) and dx=-sec(θ)tan(θ)dθ into the integral: ∫(sqrt((-sec(θ))²-1))/((-sec(θ))³) * (-sec(θ)tan(θ)dθ) Simplify the integrand: -∫tan(θ)dθ

Evaluate the integral

Evaluate the integral of -tan(θ)dθ: -∫tan(θ)dθ = -ln|sec(θ)| + C₂ where C₂ is the constant of integration.

Transform the result back into x using the inverse substitution

The inverse of the substitution x=-sec(θ) is θ=sec^(-1)(-x). Replace θ with sec^(-1)(-x) in the result: -ln|sec(sec^(-1)(-x))| + C₂ = ln|-x| + C₂ The integral for x < -1 is: ∫(sqrt(x²-1))/(x³) dx = ln|-x| + C₂ It's important to note that the two solutions have different constants of integration, C₁ and C₂, since they are derived from separate cases.

Key concepts

Trigonometric Substitution

Trigonometric substitution is a technique in calculus used to evaluate difficult integrals. This technique involves substituting a trigonometric function for a variable to simplify the integral. It is particularly helpful when dealing with expressions involving square roots. In the provided exercise, we used the substitution \( x = a \sec \theta \). Here, the function \( \sec \theta \) was chosen because it helps to simplify square roots of expressions like \( x^2 - 1 \). This simplification occurs because \( \sec^2 \theta - 1 \) equals \( \tan^2 \theta \), which eliminates the square root in the integrand.

• For \( x > 1 \), we set \( x = \sec \theta \), which means \( 0 \leq \theta < \pi/2 \).
• For \( x < -1 \), we use \( x = -\sec \theta \), meaning \( \pi/2 < \theta \leq \pi \).

Using these substitutions allows the integration of functions in terms of \( \theta \), and ultimately, the integration becomes more manageable. The new integral is often in a much simpler trigonometric form.

Definite Integration

Definite integration is the process of calculating the integral of a function with specified upper and lower bounds. It provides the net area under the curve of a function over a given interval.
In this exercise, although the focus is primarily on evaluating indefinite integrals, understanding definite integrals is crucial when the context is given further defined limits. Here, we treat the substitution as providing the necessary bounds, based on the function's behavior.
The same trigonometric substitution could be used if bounds were specified for \( x \) by determining the equivalent \( \theta \) bounds via the inverse trigonometric function. This would translate the \( x \) bounds directly to \( \theta \) bounds, ready for evaluation.

Inverse Trigonometric Functions

Inverse trigonometric functions play a key role in reversing trigonometric substitutions in integrals. Once an integral is evaluated in terms of \( \theta \), it's necessary to express the answer back in terms of \( x \). This is where inverse functions come in.
For instance, in this exercise, after evaluating the integral, the result was transformed back into the variable \( x \) by using the inverse function \( \theta = \sec^{-1}(x) \). This process is essential for providing solutions in the same variable as the original integrand. Notably, since trigonometric functions like secant and their inverses can have restrictions on their domains, it's important to consider the correct range for \( \theta \), which ensures an accurate transformation back to \( x \).

• For \( x > 1 \), \( \theta = \sec^{-1}(x) \) directly applies.
• For \( x < -1 \), use \( \theta = \sec^{-1}(-x) \) to maintain sign consistency.

Mastering how inverse trigonometric functions work provides better control over solving integrals using trigonometric substitutions.