Problem 43 - Thomas Calculus

Question

Express your answers in terms of \(x\). \(\int \sin ^{-1} x d x\)

Step-by-Step Solution

Answer

The integral is \( x \sin^{-1}(x) + \sqrt{1-x^2} + C \).

Step 1: Integration by Parts Setup

Integration by parts is defined by the formula \( \int u \, dv = uv - \int v \, du \). Let's set \( u = \sin^{-1}(x) \) and \( dv = dx \). Now, we need to compute \( du \) and \( v \).

Step 2: Compute du

To find \( du \), differentiate \( u = \sin^{-1}(x) \). The derivative is \( du = \frac{1}{\sqrt{1-x^2}} \, dx \).

Step 3: Compute v

Integrate \( dv = dx \) to find \( v \). This gives \( v = x \).

Step 4: Apply Integration by Parts

Substitute back into the integration by parts formula. We have \( \int \sin^{-1}(x) \cdot dx = x \cdot \sin^{-1}(x) - \int x \cdot \frac{1}{\sqrt{1-x^2}} \, dx \).

Step 5: Simplify and Integrate

Focus on the remaining integral, \( \int \frac{x}{\sqrt{1-x^2}} \, dx \). Use substitution by setting \( t = 1-x^2 \), then \( dt = -2x \, dx \) or \( x \, dx = -\frac{1}{2} dt \). The integral becomes \( -\frac{1}{2}\int \frac{1}{\sqrt{t}} \, dt \).

Step 6: Integrate Final Substitution

Integrate \(-\frac{1}{2} \int t^{-1/2} \, dt\). The result is \(-\frac{1}{2} \cdot 2 \sqrt{t} = -\sqrt{t} = -\sqrt{1-x^2} \) after substituting back \( t = 1-x^2 \).

Step 7: Combine Results

Put together the results of the integration by parts. The original integral evaluates to \( x \sin^{-1}(x) + \sqrt{1-x^2} + C \), where \( C \) is the constant of integration.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions help us find angles when we know the trigonometric ratios. The function \( \sin^{-1}(x) \), also known as arcsin, returns an angle whose sine is \( x \). This is particularly useful in integral calculus as it often appears in integration problems where there's a need to find antiderivatives.

• When you differentiate \( \sin^{-1}(x) \) with respect to \( x \), the derivative is \( \frac{1}{\sqrt{1-x^2}} \). This derivative is crucial because it helps set up various integration techniques like integration by parts.
• The range of \( \sin^{-1}(x) \) is between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), which means it takes values between these angles. This characteristic ensures that the solutions involving arcsin work within the principal range of the function.

Working with inverse trigonometric functions requires a good grasp of both their derivatives and ranges. This knowledge is often applied when using integration methods in calculus.

Integral Calculus

Integral calculus revolves around finding integrals, which are essentially the reverse operation of differentiation. It helps in determining areas, volumes, central points, and many other useful quantities.

• A common technique in integral calculus is integration by parts, which follows from the product rule in differentiation. It's articulated by the formula \( \int u \, dv = uv - \int v \, du \). This is particularly effective when dealing with products of functions, like an algebraic function multiplied by an inverse trigonometric function.
• Integration by parts breaks down complex integrals into simpler components. In our example \( \int \sin^{-1}(x) \, dx \), selecting the components wisely—in this case, letting \( u = \sin^{-1}(x) \) and \( dv = dx \)—allows us to solve the integral effectively.

Integral calculus is a foundational tool in mathematics and applied sciences, offering strategies like integration by parts to simplify and solve more complicated integrals.

Substitution Method

The substitution method, or integration by substitution, involves changing variables to make integration more manageable. It's analogous to the chain rule in differentiation.

• In the context of our integral, the substitution method was used to resolve \( \int \frac{x}{\sqrt{1-x^2}} \, dx \). Here, identifying \( t = 1-x^2 \) simplifies the integral. The key is to determine \( dt \) in terms of \( dx \), which is \( dt = -2x \, dx \).
• Substitution simplifies the integral to a more recognizable form, often involving simpler functions like power functions. For example, the substitution renders the integral into \( -\frac{1}{2}\int t^{-1/2} \, dt \), allowing straightforward integration.

Applying the substitution method can transform an integral into a simpler form, making it easier to solve and integrate. Knowing when and how to substitute effectively is a valuable skill in tackling a wide range of integrals.