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Magazine Chapter 8: Problem 9
Use the composite function rule to integrate the following functions: (a) \(x \sqrt{\left(1+x^{2}\right)}\) (b) \(\cos x \sin ^{3} x\) (c) \(\frac{x}{\left(1+x^{2}\right)^{2}}\) (d) \(\frac{x}{\sqrt{\left(x^{2}-1\right)}}\) (e) \(\frac{2 x+3}{x^{2}+3 x+2}\) (f) \(\sin ^{3} x \cos ^{4} x\) (g) \(\frac{x}{\left(1+x^{2}\right)^{2}}\) (h) \(\frac{x}{\sqrt{\left(4-x^{2}\right)}}\)
Short Answer
Expert verified
Use substitution and partial fractions to solve each integral problem, yielding results like \( \frac{1}{3}(1+x^2)^{3/2} + C \) and \( \ln|x+1| + \ln|x+2| + C \).
Step by step solution
01
Identify the Composite Function
For each function given, identify the inner function, which is typically inside another function or under a square root, etc. We will use substitution to simplify integration.
02
Step 2a: Integrate Part (a)
Consider the function \( x \sqrt{1+x^2} \). Use substitution: let \( u = 1+x^2 \), then \( du = 2x \, dx \), which implies \( x \, dx = \frac{1}{2} du \). Rewrite the integral as: \[ \int x \sqrt{1+x^2} \, dx = \frac{1}{2} \int \sqrt{u} \, du \] The integral becomes \( \frac{1}{2} \cdot \frac{2}{3} u^{3/2} = \frac{1}{3} u^{3/2} \). Substitute back \( u = 1+x^2 \): \[ \frac{1}{3} (1+x^2)^{3/2} + C \]
03
Step 2b: Integrate Part (b)
Consider the function \( \cos x \sin^3 x \), and set \( u = \sin x \) so that \( du = \cos x \, dx \). The integral simplifies to: \[ \int \cos x \sin^3 x \, dx = \int u^3 \, du = \frac{u^4}{4} + C \] Substitute back \( u = \sin x \): \[ \frac{1}{4} \sin^4 x + C \]
04
Step 2c: Integrate Part (c)
For \( \frac{x}{(1+x^2)^2} \), use \( u = 1+x^2 \) and \( du = 2x \, dx \), so \( x \, dx = \frac{1}{2} du \). The integral becomes \[ \int \frac{x}{(1+x^2)^2} \, dx = \frac{1}{2} \int \frac{1}{u^2} \, du \] which simplifies to \( \frac{1}{2} (-u^{-1}) = -\frac{1}{2u} + C \). Substitute back \( u = 1+x^2 \): \[ -\frac{1}{2(1+x^2)} + C \]
05
Step 2d: Integrate Part (d)
Here we have \( \frac{x}{\sqrt{x^2-1}} \). Let \( u = x^2-1 \), so \( du = 2x \, dx \), giving \( x \, dx = \frac{1}{2} du \). The integral is: \[ \int \frac{x}{\sqrt{x^2-1}} \, dx = \frac{1}{2} \int \frac{1}{\sqrt{u}} \, du \] which simplifies to \( \frac{1}{2} \cdot 2u^{1/2} = u^{1/2} + C \). Substitute back \( u = x^2-1 \): \[ \sqrt{x^2-1} + C \]
06
Step 2e: Integrate Part (e)
Consider \( \frac{2x+3}{x^2+3x+2} \), and factor the denominator as \((x+1)(x+2)\). Use partial fraction decomposition: \[ \frac{2x+3}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \] Find \( A \) and \( B \) such that: \[ 2x+3 = A(x+2) + B(x+1) \] Solving gives \( A = 1 \) and \( B = 1 \). Integrate: \[ \int \left(\frac{1}{x+1} + \frac{1}{x+2}\right) dx = \ln|x+1| + \ln|x+2| + C \]
07
Step 2f: Integrate Part (f)
For \( \sin^3 x \cos^4 x \), use the substitution \( u = \sin x \) so \( du = \cos x \, dx \). Rewrite: \[ \int \sin^3 x \cos^4 x \, dx = \int u^3 (1-u^2)^2 \, du \] Expand and integrate each term: \[ \int (u^3 - 2u^5 + u^7) \, du \] which is \[ \frac{u^4}{4} - \frac{2u^6}{6} + \frac{u^8}{8} + C \] Simplify: \[ \frac{1}{4}u^4 - \frac{1}{3}u^6 + \frac{1}{8}u^8 + C \] and substitute back \( \sin x \) for \( u \).
08
Step 2g: Integrate Part (g)
This is a duplicate of part c. The solution was already given as: \[ -\frac{1}{2(1+x^2)} + C \]
09
Step 2h: Integrate Part (h)
Consider \( \frac{x}{\sqrt{4-x^2}} \), let \( u = 4-x^2 \) and then \( du = -2x \, dx \) or \( x \, dx = -\frac{1}{2} du \). The integral is \[ \int \frac{x}{\sqrt{4-x^2}} \, dx = -\frac{1}{2} \int \frac{1}{\sqrt{u}} \, du \] which results in \( -u^{1/2} + C \). Substitute back \( 4-x^2 \) for \( u \): \[ -\sqrt{4-x^2} + C \]
10
Conclusion: Combine All Results
Each integral has been solved using substitution and potential simplification techniques. Combine each part for a complete solution:(a) \( \frac{1}{3} (1+x^2)^{3/2} + C \)(b) \( \frac{1}{4} \sin^4 x + C \)(c) \( -\frac{1}{2(1+x^2)} + C \)(d) \( \sqrt{x^2-1} + C \)(e) \( \ln|x+1| + \ln|x+2| + C \)(f) \( \frac{1}{4}\sin^4 x - \frac{1}{3}\sin^6 x + \frac{1}{8}\sin^8 x + C \)(g) \( -\frac{1}{2(1+x^2)} + C \) (same as c)(h) \( -\sqrt{4-x^2} + C \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is an important technique in integration that simplifies complex integrals. Imagine unraveling layers to find something simpler inside. Here is how it works:
- First, identify the inner function within your composite function. This is typically a part of the integrand inside another function or under a certain operation, like a square root or a power.
- Next, choose a substitution. For example, if you have \( x \sqrt{1+x^2} \), you might let \( u = 1 + x^2 \). The derivative \( du = 2x \, dx \) connects \( du \) to \( dx \), allowing you to express \( x \, dx = \frac{1}{2} \, du \).
- Substitute these variables in the original integral to transform the expression into an easier form.
- Integrate the simpler function in terms of \( u \), then revert back to the original variable by substituting \( u \) back with \( 1+x^2 \).
Partial Fraction Decomposition
Partial fraction decomposition is a key method for integrating rational functions, especially when dealing with polynomials in the denominator. Let's break it down:
- Start by factoring the denominator into linear terms or irreducible quadratic terms if needed. For example, when you have \( \frac{2x+3}{x^2+3x+2} \), the denominator factors into \((x+1)(x+2)\).
- Express the original fraction as a sum of simpler fractions, like \( \frac{A}{x+1} + \frac{B}{x+2} \), making sure to adjust the numerators according to the denominator.
- Solve for constants \( A \) and \( B \) by setting up equations based on matching coefficients from both sides of the equation.
- Integrate each simple fraction separately, which is often straightforward, such as integrating \( \frac{1}{x+1} \) to yield \( \ln|x+1| \).
Trigonometric Integration
Trigonometric integration involves functions that are products or powers of trigonometric functions. Understanding how to handle these is crucial:
- Use trigonometric identities to simplify the integrand. These might convert the integrand into a form that's easier to integrate. For instance, \( \cos x \sin^3 x \) lets you substitute \( u = \sin x \), with \( du = \cos x \, dx \), making the integral easier to solve.
- After substitution, integrate the resulting polynomial function. For example, integrate \( u^3 \) as \( \frac{u^4}{4} \).
- Ensure you substitute back the original variables after integrating, restoring the trigonometric expressions.
Integration by Parts
Integration by Parts is a useful tool derived from the product rule for differentiation. It's perfect for integrals involving the product of functions:
- This technique follows the formula: \( \int u \, dv = uv - \int v \, du \).
- Choose \( u \) and \( dv \) from your integrand such that differentiating \( u \) and integrating \( dv \) simplifies the problem. The choice of \( u \) is guided by the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential).
- Differentiate \( u \) to find \( du \) and integrate \( dv \) to get \( v \).
- Substitute into the formula and solve the resultant simpler integral.
