Problem 81 - Calculus: Graphical, Numerical, Algebraic

Question

Trigonometric Substitution Suppose \(u=\sin ^{-1} x .\) Then \(\cos u>0\) . (a) Use the substitution \(x=\sin u, d x=\cos u d u\) to show that $$\int \frac{d x}{\sqrt{1-x^{2}}}=\int 1 d u$$ (b) Evaluate \(\int 1 d u\) to show that \(\int \frac{d x}{\sqrt{1-x^{2}}}=\sin ^{-1} x+C\)

Step-by-Step Solution

Answer

The integral \(\int \frac{dx}{\sqrt{1-x^2}}\) is equal to \(\sin^{-1} x +C\).

Step 1: Initialize the substitution

We are given the substitution \(x=\sin u\) which implies \(dx=\cos u du\), so we substitute this in the given integral. The integral \(\int \frac{dx}{\sqrt{1-x^2}}\) becomes \(\int \frac{\cos u du}{\sqrt{1-\sin^2 u}}\).

Step 2: Simplify the integral

The expression in the denominator, \(\sqrt{1-\sin^2 u}\), can be simplified using Pythagorean trigonometric identity, that is \(\sqrt{1-\sin^2 u} = \cos u\). After this simplification, the integral becomes \(\int \frac{\cos u \, du}{\cos u}=\int 1 \, du\) thereby proving the first part of the problem.

Step 3: Integrate \(1 du\)

Now, we integrate \(1 du\) which gives us \(u + C\), where \(C\) is the constant of integration.

Step 4: Substitute \(u\)

Now, we substitute \(u = \sin ^{-1}x\) back into the solution, which yields \(\sin^{-1}x + C\). Hence, this shows that \(\int \frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x + C\).

Key Concepts

Integration Techniques

Integration is a powerful mathematical tool used to find areas under curves, among other applications. One important technique for solving integrals is substitution. This technique often simplifies complex expressions into more manageable ones. In the exercise above, we use trigonometric substitution to evaluate an integral.Here's how substitution works in this scenario:

• We start with the substitution of the variable. In this case, set \(x=\sin u\). This helps because it aligns with trigonometric identities that can be useful to simplify calculations.
• The differential \(dx\) gets transformed as well, resulting in \(dx = \cos u \, du\).
• These transformations turn the overall integral into a simpler form. For this problem, the integral becomes \(\int \frac{\cos u \, du}{\sqrt{1-\sin^2 u}}\).

The beauty of substitution is it turns difficult integrals into standard forms that are easier to solve. This can be particularly useful in calculus where functions often have complicated compositions.

Inverse Trigonometric Functions

Inverse trigonometric functions play a crucial role in integration, especially when dealing with integrals that naturally involve geometric interpretations. The function \(u = \sin^{-1}(x)\), pronounced as "u equals arcsine of x," is an example.Here's why inverse trigonometric functions matter:

• They help reverse the trigonometric functions, changing a ratio of sides back into an angle, which is particularly useful in integrals involving substitution.
• In this exercise, substituting \(u = \sin^{-1} x\) helps express the integral in a simpler form that corresponds directly to the inverse trigonometric functions, helping us to integrate more easily.
• Once \(u\) is obtained through integration, substituting back \(\sin^{-1} x\) ensures the solution remains in terms of the original variable, \(x\).

In many calculus problems, inverse trigonometric functions allow you to simplify the integration process, turning trigonometric identities into more straightforward expressions.

Pythagorean Identity

The Pythagorean identity is one of the fundamental relations in trigonometry that links the squares of sine and cosine functions. It states that \(\sin^2 u + \cos^2 u = 1\). This identity is pivotal when performing trigonometric substitutions in integrals.Understanding this identity is crucial because:

• It allows you to simplify square-root expressions in integrals like \(\sqrt{1 - x^2}\).
• For the exercise above, substituting \(x = \sin u\) leads to the expression \(1 - \sin^2 u\), which simplifies to \(\cos^2 u\) using the Pythagorean identity. Consequently, \(\sqrt{1 - \sin^2 u} = \cos u\).
• This simplification reduces the elements of the integral, making it straightforward to solve.

The Pythagorean identity is essential for transforming and simplifying expressions in integration, turning potentially complex calculations into elementary forms. It identifies a clear path from a trigonometric expression to an integral solution.