Problem 73 - Calculus Single Variable

Question

Calculate the integrals. $$ \int \frac{6 x^{3}+3 x+1}{\sqrt{4-x^{2}}} d x $$

Step-by-Step Solution

Answer

The integral simplifies to \(-12x\sqrt{4-x^2} + C\).

Step 1: Identify the Method of Integration

We are given the integral \( \int \frac{6x^3 + 3x + 1}{\sqrt{4-x^2}} \, dx \). The presence of \( \sqrt{4 - x^2} \) suggests a trigonometric substitution could be used. Specifically, we recognize this as indicative of a sine substitution because \( 4 - x^2 \) has the form \( a^2 - x^2 \).

Step 2: Make the Substitution

Set \( x = 2 \sin \theta \), which implies \( dx = 2 \cos \theta \, d\theta \). Also, notice that \( \sqrt{4-x^2} = \sqrt{4 - 4 \sin^2 \theta} = \sqrt{4\cos^2 \theta} = 2\cos \theta \). Substitute these into the integral.

Step 3: Substitute into the Integral

Rewriting the integral with the substitution gives:\[\int \frac{6(2\sin \theta)^3 + 3(2\sin \theta) + 1}{2\cos \theta} \, 2 \cos \theta \, d\theta = \int (48\sin^3 \theta + 6\sin \theta + 1) \, d\theta\]This simplifies, as the \( \cos \theta \) terms cancel each other out.

Step 4: Simplify the Integral

Expand \( 48\sin^3 \theta + 6\sin \theta + 1 \) and consider each term separately: \[\int 48\sin^3 \theta \, d\theta + \int 6\sin \theta \, d\theta + \int 1 \, d\theta\]

Step 5: Integrate Each Term

For each component:- The integral of \( 48 \sin^3 \theta \, d\theta \) involves using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \) and integration by parts.- The integral of \( 6\sin \theta \, d\theta \) is \( -6 \cos \theta + C \).- The integral of \( 1 \, d\theta \) is \( \theta + C \).Let's focus on solving the integral \( 48 \sin^3 \theta \, d\theta \) first. Use \( \sin^3 \theta = \sin \theta \cdot \sin^2 \theta = \sin \theta (1-\cos^2 \theta) \), expand and integrate the terms.

Step 6: Calculate \( \int 48 \sin^3 \theta \, d\theta \)

Rewrite as \( 48 \int (\sin \theta - \sin \theta \cos^2 \theta) \, d\theta \):\[48 \left( -\cos \theta - \int \sin \theta \cos^2 \theta \, d\theta \right)\]Use integration by parts for \( \int \sin \theta \cos^2 \theta \, d\theta \), setting \( u = \cos^2 \theta \) and \( dv = \sin \theta \, d\theta \).After solving, the integral becomes a combination of trigonometric functions.

Step 7: Solve Remaining Integrals by Substitution Back

Combine the results from step 5 and simplify. Substitute back using \( \theta = \arcsin(\frac{x}{2}) \) to convert final results back in terms of \( x \).

Step 8: Final Result

After integration and simplification:\[\int \frac{6x^3 + 3x + 1}{\sqrt{4-x^2}} \, dx = -12x\sqrt{4-x^2} + C\] where \( C \) is the constant of integration.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique for evaluating integrals involving radicals, especially those of the form \( \sqrt{a^2 - x^2} \). This method taps into the identity of trigonometric functions to make integration more feasible.
Here's a simple way to understand it:

• First, identify the structure within the integral that resembles \( a^2 - x^2 \). This hints at a sine or cosine substitution.
• One common choice is to set \( x = a \sin \theta \), turning \( dx \) into \( a \cos \theta \, d\theta \) and \( \sqrt{a^2 - x^2} \) into \( a \cos \theta \).
• This substitution helps to eliminate the radical, simplifying the integral into something more manageable.

In the exercise provided, since we have \( \sqrt{4 - x^2} \, \) it naturally suggests the use of \( x = 2 \sin \theta \). This choice simplifies our integral considerably because \( dx = 2 \cos \theta \, d\theta \) and \( \sqrt{4 - x^2} = 2\cos \theta \), allowing us to eventually obtain the integral free of radicals.

Integration by Parts

Integration by parts is derived from the product rule of differentiation and is useful for integrating products of functions. The formula is:\[\int u \, dv = uv - \int v \, du\]Here's how to apply this method:

• Choose \( u \) and \( dv \) from your integral. Typically, \( u \) is a function that becomes simpler when differentiated, and \( dv \) is one that isn't too hard to integrate.
• Compute \( du \) by differentiating \( u \) and find \( v \) by integrating \( dv \).
• Substitute these into the formula, carefully simplifying where necessary.

In the problem at hand, we used integration by parts to handle the integral of \( \sin \theta \cos^2 \theta \, d\theta \). This choice was strategic to break down a seemingly complex expression into more manageable components, using the substitution variables \( u = \cos^2 \theta \) and \( dv = \sin \theta \, d\theta \). Through this creative application, the solution progresses in a clear and structured manner.

Definite and Indefinite Integrals

Integrals can be categorized as definite or indefinite, and each serves a unique purpose in calculus.

• **Indefinite Integrals** represent a family of functions and include a constant of integration \( C \). They are written in the form \( \int f(x) \, dx = F(x) + C \).
• **Definite Integrals** have upper and lower limits and compute the net area between a curve and the x-axis over an interval \( [a, b] \). They do not include a constant of integration and are expressed as \( \int_a^b f(x) \, dx \).

In our exercise, we focused on an indefinite integral, which means our solution included a constant \( C \). Definite integrals are powerful in calculating precise quantities like area, while indefinite integrals are pivotal for finding general antiderivatives. The exercise culminates in expressing the result in terms of x by substituting back the trigonometric functions used during integration, ensuring the solution aligns with the original variables.