Problem 82 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int \frac{e^{\cos ^{-1} x} d x}{\sqrt{1-x^{2}}}$$

Step-by-Step Solution

Answer

The integral evaluates to \(-e^{\cos^{-1}(x)} + C\).

Step 1: Recognize the substitution

Notice that the term \( \sqrt{1-x^2} \) appears in the denominator. This suggests a substitution involving \( \cos^{-1}(x) \). Let \( u = \cos^{-1}(x) \), which implies that \( x = \cos(u) \) and \( dx = -\sin(u) \, du \).

Step 2: Substitute the appropriate expressions

Substitute \( x = \cos(u) \) and \( dx = -\sin(u) \, du \) into the integral. This yields:\[ \int \frac{e^{u} (-\sin(u)) \, du}{\sqrt{1 - \cos^2(u)}} \]The expression \( \sqrt{1 - \cos^2(u)} \) simplifies to \( \sin(u) \).

Step 3: Simplify resulting integral

After substitution, the integral simplifies to:\[ \int e^{u} (-\sin(u)) \, du \frac{1}{\sin(u)} \] This further simplifies to:\[ \int -e^{u} \, du \] which is equivalent to:\[ -\int e^{u} \, du \]

Step 4: Integrate with respect to u

The integral \(-\int e^{u} \, du\) can be directly determined as \(-e^{u} + C\), where \( C \) is the constant of integration.

Step 5: Substitute back for x and conclude

Since we made the substitution \( u = \cos^{-1}(x) \), substitute this back into the solution:\[ -e^{\cos^{-1}(x)} + C \]This gives us the final expression for the integral in terms of \( x \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique used in calculus for simplifying integrals containing square roots of quadratic expressions. It often involves substituting a trigonometric function in place of a variable, based on common trigonometric identities.
In this exercise, we view the expression \( \sqrt{1-x^2} \). This is reminiscent of the Pythagorean identity \( \sin^2(u) + \cos^2(u) = 1 \). Here, substituting \( x = \cos(u) \) transforms \( \sqrt{1-x^2} \) into \( \sin(u) \), making the integral simpler.

Here’s why this works well:

• It transforms the square root expression into a clean trigonometric function.
• It can turn an awkward integral into a basic one, which is easier to evaluate.
• It relies heavily on the identities and derivatives of trigonometric functions.

Thus, this substitution enables us to efficiently compute an otherwise challenging integral.

Definite Integrals

Definite integrals are a staple in calculus that enable the computation of the area under curves. They are typically denoted with bounds, though our integral was indefinite.
To solve definite integrals, we often use the Fundamental Theorem of Calculus, which links differentiation and integration in a powerful relationship.

In the context of this problem:

• We solve without bounds, but imagine having them, this would find the net area under the curve between those points.
• The substitution simplifies the integrand first, making it easier to work with should you apply definite bounds later.
• Transformations in definite integrals maintain function equality, allowing shifts to new variable bounds after substitution.

The integration process here culminated in the antiderivative \(-e^{u} + C\). This would be adapted to include specific limits if we moved to definite integrals.

Arc Cosine Function

The arc cosine function, \( \cos^{-1}(x) \), is the inverse of the cosine function. It returns the angle whose cosine is \( x \). Given a transformed interval from \([-1, 1]\), arc cosine is often applied when dealing with circular trigonometric problems.
This function plays a crucial role in the substitution process. Here’s how:

• In substitution, \( u = \cos^{-1}(x) \) is key because it facilitates the transformation of the integral.
• This allows converting \( x = \cos(u) \), resulting in \( dx = -\sin(u) \, du \), a necessity for simplifying the integral.
• It provides the necessary relationship between variables to use trigonometric identities effectively, as seen in this problem.

Ultimately, the arc cosine enables the seamless transition from a polynomial form to a trigonometric one, simplifying the integration process.