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Integral Calculus Problem:
[tex]\int\limits^1_4 {\frac{cosx}{1+e^1/x} } \, dx[/tex] = _

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Answer:

[tex]\displaystyle \int\limits^1_4 {\frac{\cos x}{1 + \frac{e}{x}}} \, dx \approx \boxed{0.862854}[/tex]

General Formulas and Concepts:

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Addition/Subtraction]:
[tex]\displaystyle (u + v)' = u' + v'[/tex]

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals

Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

Integration Methods: U-Substitution and U-Solve

Special Integrals:

  • Sine Integral:
    [tex]\displaystyle \int {\frac{\sin x}{x}} \, dx = \text{Si} (u) + C[/tex]
  • Cosine Integral:
    [tex]\displaystyle \int {\frac{\cos x}{x}} \, dx = \text{Ci} (u) + C[/tex]

Step-by-step explanation:

*Note:
The problem is too big to fit all work. I will assume that you know how to do basic calculus.

Step 1: Define

Identify given.

[tex]\displaystyle \int\limits^1_4 {\frac{\cos x}{1 + \frac{e}{x}}} \, dx[/tex]

Step 2: Integrate Pt. 1

  1. [Integrand] Rewrite:
    [tex]\displaystyle \begin{aligned}\int\limits^1_4 {\frac{\cos x}{1 + \frac{e}{x}}} \, dx & = \int\limits^1_4 {\frac{x \cos x}{x + e} \, dx \leftarrow \\\end{aligned}[/tex]

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

  1. Set u:
    [tex]\displaystyle u = x + e[/tex]
  2. [u] Differentiate [Derivative Rules and Properties]:
    [tex]\displaystyle du = dx[/tex]

Step 4: Integrate Pt. 3

  1. [Integral] Apply Integration Method [U-Substitution]:
    [tex]\displaystyle \begin{aligned}\int\limits^1_4 {\frac{\cos x}{1 + \frac{e}{x}}} \, dx & = \int\limits^1_4 {\frac{x \cos x}{x + e} \, dx \\& = \int\limits^{x = 1}_{x = 4} {\frac{(u - e) \cos (u - e)}{u} \, du \leftarrow \\\end{aligned}[/tex]
  2. [Integral] Rewrite [Integration Property - Addition/Subtraction]:
    [tex]\displaystyle \begin{aligned}\int\limits^1_4 {\frac{\cos x}{1 + \frac{e}{x}}} \, dx & = \int\limits^1_4 {\frac{x \cos x}{x + e} \, dx \\& = \int\limits^{x = 1}_{x = 4} {\frac{(u - e) \cos (u - e)}{u} \, du \\ & = \int\limits^{x = 1}_{x = 4} {\cos ( u - e)} \, du - \int\limits^{x = 1}_{x = 4} {\frac{e \cos (u - e)}{u}} \, du \leftarrow \\\end{aligned}[/tex]
  3. [2nd Integral] Rewrite [Integration Property - Multiplied Constant]:
    [tex]\displaystyle \begin{aligned}\int\limits^1_4 {\frac{\cos x}{1 + \frac{e}{x}}} \, dx & = \int\limits^{x = 1}_{x = 4} {\cos ( u - e)} \, du - e \int\limits^{x = 1}_{x = 4} {\frac{\cos (u - e)}{u}} \, du \leftarrow \\\end{aligned}[/tex]

Step 5: Integrate Pt. 4

Identify variables for u-substitution for the 1st integral.

Use another variable besides u to avoid confusion with earlier substitutions.

  1. Set v:
    [tex]\displaystyle v = u - e[/tex]
  2. [v] Differentiate:
    [tex]\displaystyle dv = du[/tex]

Step 6: Integrate Pt. 5

Solve the 1st integral using basic integration techniques listed under "Calculus":
[tex]\displaystyle \begin{aligned}\int\limits^{x = 1}_{x = 4} {\cos (u - e)} \, du & = \int\limits^{x = 1}_{x = 4} {\cos v} \, dv \\& = \sin v \bigg| \limits^{x = 1}_{x = 4} \\& = \sin (u - e) \bigg| \limits^{x = 1}_{x = 4} \\& = \sin (x) \bigg| \limits^{x = 1}_{x = 4} \\& = \sin 1 - \sin 4\end{aligned}[/tex]

Step 7: Integrate Pt. 6

To solve the 2nd integral, we use the same methods as "Step 6":

[tex]\displaystyle\begin{aligned}\int\limits^{x = 1}_{x = 4} {\frac{\cos (u - e)}{u}} \, du & = \int\limits^{x = 1}_{x = 4} {\frac{\sin e \sin u + \cos e \cos u}{u}} \, du \\& = \sin e \int\limits^{x = 1}_{x = 4} {\frac{\sin u}{u}} \, du + \cos e \int\limits^{x = 1}_{x = 4} {\frac{\cos u}{u}} \, du \\& = \sin (e) \text{Si} (u) \bigg| \limits^{x = 1}_{x = 4} + \cos (e) \text{Ci} (u) \bigg| \limits^{x = 1}_{x = 4} \\\end{aligned}[/tex]

[tex]\displaystyle\begin{aligned}\int\limits^{x = 1}_{x = 4} {\frac{\cos (u - e)}{u}} \, du & = \sin (e) \text{Si} (x + e) \bigg| \limits^{x = 1}_{x = 4} + \cos (e) \text{Ci} (x + e) \bigg| \limits^{x = 1}_{x = 4} \\& = \bigg[ \sin (e) \text{Si} (1 + e) - \sin (e) \text{Si} (4 + e) \bigg] + \bigg[ \cos (e) \text{Ci} (1 + e) - \cos (e) \text{Ci} (4 + e) \bigg] \\\end{aligned}[/tex]

Step 8: Integrate Pt. 7

Combine our 2 integral values to obtain a final answer:

[tex]\displaystyle \int\limits^1_4 {\frac{\cos x}{1 + \frac{e}{x}}} \, dx & = \int\limits^{x = 1}_{x = 4} {\cos ( u - e)} \, du - e \int\limits^{x = 1}_{x = 4} {\frac{\cos (u - e)}{u}} \, du \\\begin{aligned}& = \sin 1 - \sin 4 - e \bigg[ \sin (e) \text{Si} (1 + e) - \sin (e) \text{Si} (4 + e) + \cos (e) \text{Ci} (1 + e) - \cos (e) \text{Ci} (4 + e) \bigg] \\& \approx \boxed{0862854} \\\end{aligned}[/tex]

∴ we have evaluated the given special definite integral.

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Topic: Calculus

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