Evaluate The Integral. 0^ 3x^3 Cos(x)dx
When it comes to calculus, one of the fundamental tasks is evaluating definite integrals, which are essential for understanding areas, accumulated quantities, and many applications across science and engineering. The integral in question, from 0 to 3, of the function \( 3x^3 \cos(x) \, dx \), presents a classic example of an integral that involves polynomial and trigonometric functions combined. This article provides a comprehensive guide to evaluating this integral, exploring various methods including integration by parts, and offering detailed step-by-step solutions to enhance your calculus skills.
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Understanding the Integral \( \int_0^3 3x^3 \cos(x) \, dx \)
Before diving into calculations, it is important to understand the structure of the integral and the properties of the functions involved.
Components of the Integral
The integrand \( 3x^3 \cos(x) \) consists of:
- A polynomial part: \( 3x^3 \)
- A trigonometric part: \( \cos(x) \)
The multiplication of these two functions suggests that the integral may require techniques like integration by parts, particularly because neither polynomial nor cosine alone simplifies the other directly.
Importance of Properly Setting Up the Integral
Correctly setting up the integral involves:
- Recognizing the limits of integration: from 0 to 3
- Identifying the integrand's structure
- Deciding on the appropriate method for evaluation
Since the integrand is a product of polynomial and cosine functions, integration by parts is a suitable method.
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Methodology for Evaluating \( \int_0^3 3x^3 \cos(x) \, dx \)
The integral can be tackled using the integration by parts technique, which is based on the formula:
\[
\int u \, dv = uv - \int v \, du
\]
Choosing the right parts \( u \) and \( dv \) is crucial for simplifying the integral.
Step 1: Selecting \( u \) and \( dv \)
For the integral \( \int 3x^3 \cos(x) \, dx \), a good choice is:
- \( u = 3x^3 \) (since polynomial reduces upon differentiation)
- \( dv = \cos(x) \, dx \)
This choice leverages the power of polynomial differentiation to simplify the integral.
Step 2: Computing \( du \) and \( v \)
Calculations:
- \( du = 9x^2 \, dx \)
- \( v = \int \cos(x) \, dx = \sin(x) \)
Step 3: Applying the Integration by Parts Formula
Applying the formula:
\[
\int 3x^3 \cos(x) \, dx = 3x^3 \sin(x) - \int \sin(x) \cdot 9x^2 \, dx
\]
Simplify:
\[
= 3x^3 \sin(x) - 9 \int x^2 \sin(x) \, dx
\]
Now, the problem reduces to evaluating \( \int x^2 \sin(x) \, dx \), which again involves polynomial and trigonometric functions.
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Evaluating \( \int x^2 \sin(x) \, dx \) via Integration by Parts
This integral requires a second application of integration by parts.
Step 1: Choosing \( u \) and \( dv \)
- \( u = x^2 \)
- \( dv = \sin(x) \, dx \)
- \( du = 2x \, dx \)
- \( v = -\cos(x) \)
Step 2: Applying the formula
\[
\int x^2 \sin(x) \, dx = -x^2 \cos(x) - \int -\cos(x) \cdot 2x \, dx
\]
Simplify:
\[
= -x^2 \cos(x) + 2 \int x \cos(x) \, dx
\]
Now, evaluate \( \int x \cos(x) \, dx \) using integration by parts again.
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Evaluating \( \int x \cos(x) \, dx \) for Complete Solution
Applying integration by parts:
- \( u = x \)
- \( dv = \cos(x) \, dx \)
Calculations:
- \( du = dx \)
- \( v = \sin(x) \)
Applying the formula:
\[
\int x \cos(x) \, dx = x \sin(x) - \int \sin(x) \, dx
\]
\[
= x \sin(x) + \cos(x) + C
\]
Putting it all together, the integral \( \int x^2 \sin(x) \, dx \) becomes:
\[
= -x^2 \cos(x) + 2 \left( x \sin(x) + \cos(x) \right) + C
\]
Simplify:
\[
= -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) + C
\]
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Final Assembly of the Integral \( \int 3x^3 \cos(x) \, dx \)
Recall the earlier expression:
\[
\int 3x^3 \cos(x) \, dx = 3x^3 \sin(x) - 9 \int x^2 \sin(x) \, dx
\]
Substitute the evaluated \( \int x^2 \sin(x) \, dx \):
\[
= 3x^3 \sin(x) - 9 \left( -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) \right) + C
\]
Distribute:
\[
= 3x^3 \sin(x) + 9x^2 \cos(x) - 18x \sin(x) - 18 \cos(x) + C
\]
This is the indefinite integral. To evaluate the definite integral from 0 to 3, substitute the upper and lower limits and subtract.
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Evaluating the Definite Integral from 0 to 3
Calculate at \( x = 3 \):
\[
I(3) = 3(3)^3 \sin(3) + 9(3)^2 \cos(3) - 18(3) \sin(3) - 18 \cos(3)
\]
Calculate at \( x = 0 \):
\[
I(0) = 3(0)^3 \sin(0) + 9(0)^2 \cos(0) - 18(0) \sin(0) - 18 \cos(0)
\]
Simplify:
- \( I(0) = 0 + 0 - 0 - 18 \times 1 = -18 \)
Now, substitute \( x = 3 \):
- \( 3 \times 27 \times \sin(3) = 81 \sin(3) \)
- \( 9 \times 9 \times \cos(3) = 81 \cos(3) \)
- \( -18 \times 3 \times \sin(3) = -54 \sin(3) \)
- \( -18 \times \cos(3) = -18 \cos(3) \)
Adding these:
\[
I(3) = 81 \sin(3) + 81 \cos(3) - 54 \sin(3) - 18 \cos(3)
\]
Combine like terms:
\[
= (81 \sin(3) - 54 \sin(3)) + (81 \cos(3) - 18 \cos(3))
\]
\[
= 27 \sin(3) + 63 \cos(3)
\]
Finally, the definite integral:
\[
\boxed{
\int_0^3 3x^3 \cos(x) \, dx = I(3) - I(0) = (27 \sin(3) + 63 \cos(3)) - (-18) = 27 \sin(3) + 63 \cos(3) + 18
}
\]
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Conclusion and Summary
Evaluating the integral \( \int_0^3 3x^3 \cos(x) \, dx \) involves multiple steps of integration by parts, applied recursively to handle the polynomial and trigonometric components. Here is a summarized process:
- Use integration by parts to reduce the polynomial degree.
- Continue applying integration by parts until the polynomial term is fully differentiated.
- Carefully substitute limits to evaluate the definite integral.
- Combine like terms to arrive at the final value, expressed in terms of sine and cosine functions at the upper limit, along with a constant.
This process exemplifies the power of integration techniques in calculus, especially when dealing with complex polynomial-trigonometric integrals.
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