Problem 33 - Calculus for Biology and Medicine

Question

Evaluating the integral \(\int \arcsin x d x\) requires two steps. (a) Write $$ \arcsin x=1 \cdot \arcsin x $$ and integrate by parts once to show that $$ \int \arcsin x d x=x \arcsin x-\int \frac{x}{\sqrt{1-x^{2}}} d x $$ (b) Use substitution to compute $$ \int \frac{x}{\sqrt{1-x^{2}}} d x $$ and combine your result in (a) with (7.11) to complete the computation of \(\int \arcsin x d x\).

Step-by-Step Solution

Answer

\( \int \arcsin x \, dx = x \arcsin x + \sqrt{1-x^2} + C \).

Step 1: Set Up Integration by Parts

We begin the integration by parts by setting \( u = \arcsin x \) and \( dv = dx \). Now, differentiate \( u \) to get \( du = \frac{1}{\sqrt{1-x^2}} dx \), and integrate \( dv \) to obtain \( v = x \).

Step 2: Apply Integration by Parts Formula

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Substitute into this formula: \[ \int \arcsin x \, dx = x \arcsin x - \int x \frac{1}{\sqrt{1-x^2}} \, dx \].

Step 3: Set Up Substitution for Remaining Integral

In the integral \( \int \frac{x}{\sqrt{1-x^2}} dx \), use the substitution \( t = 1 - x^2 \), so \( dt = -2x dx \). Rearrange to get \( x dx = -\frac{1}{2} dt \).

Step 4: Substitute and Simplify the Integral

Substitute and simplify the integral: \[ \int \frac{x}{\sqrt{1-x^2}} dx = \int \frac{-1/2}{\sqrt{t}} dt = -\frac{1}{2} \int t^{-1/2} dt \].

Step 5: Integrate and Back-Substitute

Integrate \( -\frac{1}{2} \int t^{-1/2} dt \) to get \( -\sqrt{t} + C \). Substitute back for \( t = 1-x^2 \): \[ -\sqrt{1-x^2} + C \].

Step 6: Combine Results

Combine the results from parts (a) and (b) to get the final integral: \[ \int \arcsin x \, dx = x \arcsin x + \sqrt{1-x^2} + C \].

Key Concepts

Definite Integrals

A definite integral is a fundamental concept in calculus, representing the area under a curve on a particular interval. This type of integral is marked by specific upper and lower bounds. When calculating a definite integral, you compute the integral of a function between these bounds. For example, if you're given a function \( f(x) \), the definite integral from \( a \) to \( b \) is represented as: \[ \int_{a}^{b} f(x) \, dx \] Here’s the process often involves:

• Finding the antiderivative \( F(x) \) of \( f(x) \).
• Evaluating \( F(x) \) at the upper and lower limits \( b \) and \( a \).
• Calculating \( F(b) - F(a) \) to find the exact area.

In the context of indefinite integrals like \( \int \arcsin x \, dx \), we are technically dealing with indefinite integrals, though understanding its definite counterpart is crucial to grasp the whole picture of integration. A definite integral concerns actual areas and real-world applications where bounds are essential.

Substitution Method

The substitution method is like swapping parts of a problem for ones that are easier to handle. This method is especially helpful in evaluating integrals, particularly when they involve composite functions. Essentially, you substitute a part of the integrand with a new variable, simplifying the integration process. ### How it WorksImagine an integral such as \( \int f(g(x))g'(x) dx \). You can make it easier by:

• Setting \( u = g(x) \).
• Then, \( du = g'(x) dx \).
• The integral now transforms into \( \int f(u) \, du \).

### Application in Our ExampleIn the exercise involving \( \int \frac{x}{\sqrt{1-x^{2}}} dx \), substituting simplifies the difficult expression, allowing for easier integration:

• Start with \( t = 1-x^{2} \) which leads to \( dt = -2x dx \), simplifying the original expression.
• You can write \( x dx \) as \( -\frac{1}{2} dt \) making it easier to integrate.

Substitution is a powerful technique that reduces complexity, making tough integrals manageable. After substitution, integrate with respect to the new variable, then back-substitute to find the solution in terms of the original variable.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, play a significant role in calculus, particularly in integration. They relate angles in a triangle to the ratios of its sides. These functions show up frequently in integrals due to their cyclic nature and presence in many physical phenomena.### Understanding \( \arcsin \)The \( \arcsin \) function, also known as the inverse sine, is central to the exercise given. If you start with \( \sin(y) = x \), then \( \arcsin(x) = y \). That means \( \arcsin \) gives you the angle whose sine is \( x \). When integrating \( \int \arcsin x \, dx \), you're finding the area under the curve described by this function.### Why Trigonometric Functions?

• They simplify during integration and differentiation.
• Help in transforming seemingly complex integrals into simpler forms.
• Often allow use of clever substitutions and identities to break down problems.

Integrals involving trigonometric functions can often be solved through methods like substitution or integration by parts, just like in this exercise. Trigonometric functions offer unique perspectives to solving integral problems, transforming difficulty into simplicity.