Problem 6
Question
Use integration by parts to evaluate the integrals. $$ \int x \sin (1-2 x) d x $$
Step-by-Step Solution
Verified Answer
\( \int x \sin(1-2x) \, dx = \frac{1}{2} x \cos(1-2x) + \frac{1}{4} \sin(1-2x) + C \)
1Step 1: 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 \( dv = \sin(1-2x) \, dx \).
2Step 2: 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 \( v = \int \sin(w) \frac{-1}{2} \, dw = \frac{1}{2} \cos(w) = \frac{1}{2} \cos(1-2x) \).
3Step 3: 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 \( x \cdot \frac{1}{2} \cos(1-2x) - \int \frac{1}{2} \cos(1-2x) \, dx \).
4Step 4: 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 \( \int \frac{1}{2} \cos(w) \frac{-1}{2} \, dw = -\frac{1}{4} \sin(w) = -\frac{1}{4} \sin(1-2x) \).
5Step 5: 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.
Key Concepts
DifferentiationSubstitution MethodDefinite Integrals
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 \( du = dx \).
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:
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.
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.
Other exercises in this chapter
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