Problem 8 - Thomas Calculus

Question

Evaluate the integrals. \(\int \sin ^{-1} y d y\)

Step-by-Step Solution

Answer

\( \int \sin^{-1} y \, dy = y \sin^{-1} y + \sqrt{1-y^2} + C \).

Step 1: Understand the Problem

We need to evaluate the integral \( \int \sin^{-1} y \, dy \). This function is the inverse sine function and it's known as arc sine or \( \arcsin(y) \). The approach that often works well for such integrals is integration by parts.

Step 2: Choose Functions for Integration by Parts

For integration by parts, we use the formula: \[ \int u \, dv = uv - \int v \, du \].We need to choose \( u = \sin^{-1} y \) and \( dv = dy \).Then, find \( du \) and integrate \( dv \).

Step 3: Differentiate and Integrate

Differentiate \( u = \sin^{-1} y \) to get \( du = \frac{1}{\sqrt{1-y^2}} \, dy \).Integrate \( dv = dy \) to find \( v = y \).

Step 4: Apply Integration by Parts

We substitute into the integration by parts formula using our choices:\[ \int \sin^{-1} y \, dy = y \sin^{-1} y - \int y \cdot \frac{1}{\sqrt{1-y^2}} \, dy \].

Step 5: Simplify the Remaining Integral

Focus on the remaining integral: \( \int \frac{y}{\sqrt{1-y^2}} \, dy \).Recognize this as a standard form and solve it using substitution: let \( t = 1-y^2 \), so \( dt = -2y \, dy \), giving \( -\frac{1}{2} \int \frac{1}{\sqrt{t}} \, dt \).

Step 6: Solve the Substitution

The integral \( -\frac{1}{2} \int \frac{1}{\sqrt{t}} \, dt \) becomes \(-\sqrt{t} + C\).Substitute back to the original variable: \( -\sqrt{1-y^2} + C \).

Step 7: Combine the Results

Combine all parts from the integration by parts:\[ y \sin^{-1} y - (-\sqrt{1-y^2}) + C = y \sin^{-1} y + \sqrt{1-y^2} + C \].

Key Concepts

Inverse Trigonometric Functions

The function \( \sin^{-1} y \), also known as \( \arcsin(y) \), is an example of an inverse trigonometric function. Inverse trigonometric functions are used to find the angles when the value of the trigonometric ratio is known.
It's important to understand that for any function \( \sin(x) \), its inverse \( \sin^{-1}(y) \) will take the output of \( \sin \), which is \( y \), and return the corresponding angle \( x \) in specified ranges.
For the sine function:

• Its range is between \(-1\) and \(1\).
• The inverse function is defined within \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) radians.

These features make inverse trigonometric functions essential in various mathematical applications, including calculus and geometry, especially when dealing with angles and periodic phenomena.

Definite and Indefinite Integrals

In calculus, integrals are fundamental concepts. They represent accumulations of quantities, such as areas under curves, collectively termed as definite and indefinite integrals.
An indefinite integral, such as \( \int \sin^{-1} y \, dy \), involves finding an antiderivative. This essentially means reversing differentiation to discover the function that was differentiated.
Definite integrals, on the other hand, compute numerical values over a defined interval. They evaluate the total accumulation, area, or net change over a specific domain.
Key differences include:

• Indefinite integrals include a constant \( C \), representing any possible vertical shift since constants vanish upon differentiation.
• Definite integrals don't include \( C \) since they result in specific values dependent on the bounds provided.

Both concepts are crucial in solving practical problems involving continuous change and are extensively supported by a rich set of rules in calculus.

Substitution Method

The substitution method is a technique used in integration to simplify complex integrals. It involves substituting part of the integral expression with a new variable, making the expression easier to integrate.
In the solution of \( \int \sin^{-1} y \, dy \), after applying integration by parts, the remaining integral \( \int \frac{y}{\sqrt{1-y^2}} \, dy \) can be approached using substitution.

• We set \( t = 1 - y^2 \), which transforms our integral to a simpler form.
• This change allows us to use a practical formula \( \int \frac{1}{\sqrt{t}} \, dt \), easier to solve analytically.

Substitution is effective, especially when faced with integrands containing composite functions, allowing integration to continue seamlessly, eventually transforming back to the original variable. This method is similar, in essence, to the chain rule used in differentiation, enabling a reverse approach to surface what was obscured in the original integrand.