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Chapter 7: Problem 6

Use integration by parts to evaluate the integrals. $$ \int x \sin (1-2 x) d x $$

Short Answer

Expert verified
\( \int x \sin(1-2x) \, dx = \frac{1}{2} x \cos(1-2x) + \frac{1}{4} \sin(1-2x) + C \)

Step by step solution

01

Identify Functions for Integration by Parts

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). We need to identify \( u \) and \( dv \) from the integral \( \int x \sin(1-2x) \, dx \). Choose \( u = x \) and \( dv = \sin(1-2x) \, dx \).
02

Differentiate and Integrate Functions

Differentiate \( u = x \) to get \( du = dx \). Now, integrate \( dv = \sin(1-2x) \, dx \). Use substitution: let \( w = 1-2x \), then \( dw = -2 \, dx \), resulting in \( v = \int \sin(w) \frac{-1}{2} \, dw = \frac{1}{2} \cos(w) = \frac{1}{2} \cos(1-2x) \).
03

Apply the Integration by Parts Formula

Substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula: \( \int x \sin(1-2x) \, dx = uv - \int v \, du \). This gives \( x \cdot \frac{1}{2} \cos(1-2x) - \int \frac{1}{2} \cos(1-2x) \, dx \).
04

Evaluate the Remaining Integral

Focus on \( \int \frac{1}{2} \cos(1-2x) \, dx \). Use substitution: let \( w = 1-2x \), then \( dw = -2 \, dx \), resulting in \( \int \frac{1}{2} \cos(w) \frac{-1}{2} \, dw = -\frac{1}{4} \sin(w) = -\frac{1}{4} \sin(1-2x) \).
05

Write the Final Solution

Substitute the evaluated integral back, obtaining: \( \frac{1}{2} x \cos(1-2x) + \frac{1}{4} \sin(1-2x) + C \), where \( C \) is the constant of integration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a fundamental concept in calculus, primarily focusing on finding the rate of change of a function with respect to a variable. In simpler words, it tells us how a function behaves as its input changes. For instance, in our problem, the function given is a simple linear term, namely, \( u = x \).
When you differentiate \( u = x \), you get \( du = dx \).
  • The differentiation of a constant is zero, hence if you had a constant instead of \( x \), its differentiation would simply be zero.
  • Differentiation helps in finding slopes of tangent lines to the curve of a function at any given point.
      • Thus, differentiation here aids in simplifying parts of the integrand, making it easier to evaluate through integration by parts.
Substitution Method
Sometimes, integration isn't straightforward and requires a simpler approach to solve. This is where the substitution method is a valuable technique. Substitution rearranges the integral to make it easier to solve by introducing a new variable, like in our exercise.
As shown:
  • For \( dv = \sin(1-2x) \), substituting \( w = 1-2x \) changes the variable, making differentiation of \( w \) simpler as \( dw = -2dx \).
  • This substitution transforms the integral into a friendlier form that can be easily manipulated and evaluated.
  • Eventually, the integral \( \int \sin(w) \frac{-1}{2} \, dw \) becomes feasible to solve using basic trigonometric integration.
Ultimately, the substitution method changes complex expressions into simpler forms, streamlining the integration process.
Definite Integrals
While our problem focuses on indefinite integrals, definite integrals have a specific range or limits, usually denoting the area under a curve on a particular interval. However, understanding them is crucial while dealing with integration.
  • Definite integrals result in a specific number rather than a function plus a constant \( C \). This number represents the total accumulation of the value represented by the function over a given interval.
  • Conceptually, they are short-hand for finding the exact sum provided by a function over specified bounds, rather than calculating it bit-by-bit.
  • Definite integrals often utilize the evaluation theorem, involving plugging upper and lower limits back into the solution of the indefinite integral.
Even though definite integrals differ slightly from the integration by parts exercise, understanding their reasoning lays the groundwork for more comprehensive applications of integration in calculus.

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