Problem 4

Question

Use integration by parts to evaluate the integrals. $$ \int 3 x \cos (4-x) d x $$

Step-by-Step Solution

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Answer

The integral evaluates to $-3x \sin(4-x) - 3 \cos(4-x) + C$.

1Step 1: Choose u and dv

In integration by parts, we start by choosing parts of the integrand to be $ u $ and $ dv $. Let $ u = 3x $, so that $ du = 3 \, dx $. Then, let $ dv = \cos(4-x) \, dx $.

2Step 2: Find v by integrating dv

We need to find $ v $ by integrating $ dv $. We know $ dv = \cos(4-x) \, dx $. First, perform a substitution with $ u = 4-x $, yielding $ du = -dx $. This transforms the integral $ \int \cos(4-x) \, dx = -\int \cos u \, du = -\sin(u) + C $. Substituting back gives $ v = -\sin(4-x) $.

3Step 3: Apply the Integration by Parts Formula

The integration by parts formula is $ \int u \, dv = uv - \int v \, du $. Substituting our parts, we have $ \int 3x \cos(4-x) \, dx = 3x(-\sin(4-x)) - \int -\sin(4-x) \, 3 \, dx $. This simplifies to $ -3x \sin(4-x) + 3 \int \sin(4-x) \, dx $.

4Step 4: Solve the Remaining Integral

We already have part of the solution: $ -3x \sin(4-x) $. Now, let's find $ 3 \int \sin(4-x) \, dx $. Use substitution again: let $ u = 4-x $ $ du = -dx $, the integral becomes $ 3(-\cos(u)) = -3\cos(u) + C $. Substitute back to obtain $ -3 \cos(4-x) $.

5Step 5: Combine the Parts

Combine both parts to get the final integration result: $-3x \sin(4-x) + 3(-\cos(4-x))$. This simplifies to $-3x \sin(4-x) - 3 \cos(4-x) + C$ where $ C $ is an arbitrary constant.

Key Concepts

Integral CalculusSubstitution MethodTrigonometric Integration

Integral Calculus

Integral calculus is a fundamental branch of mathematics that focuses on the concept of integration. It allows us to find the total accumulation of quantities or the area under curves, among other applications. In the exercise provided, we deal with the integration of a function using a technique known as integration by parts, which is a key concept in integral calculus.

Integration by parts is derived from the product rule for differentiation and is expressed as: \[\int u \, dv = uv - \int v \, du\]

Here, $ u $ and $ dv $ are parts of the original function $ f(x) $ being integrated.
Choosing the correct $ u $ and $ dv $ is crucial to simplifying the integral.

Understanding how to utilize integration by parts within integral calculus can simplify solving complex integrals and is therefore an essential skill. In our example, the integral of $ 3x \cos(4-x) $ demonstrates how we can break the problem down into simpler parts by smartly choosing $ u $ and $ dv $.

Substitution Method

The substitution method is another powerful and often complementary integration technique used in calculus. It helps to simplify integrals by changing the variable of integration.

In the provided example, substitution plays a key role in evaluating the integral of the function $ \cos(4-x) $. Here's how this works:

We substitute $ u = 4-x $, leading to $ du = -dx $.
This changes the integral into a form easier to handle, i.e., $ \int \cos(4-x) \, dx = -\int \cos(u) \, du $.

Integrating $ \cos(u) $ gives $ -\sin(u) $, and substituting back $ u = 4-x $ results in $ v = -\sin(4-x) $. This substitution streamlines the computation, allowing us to focus on solving more straightforward parts of the problem.

Trigonometric Integration

Trigonometric integration involves integrating expressions that contain trigonometric functions. This process requires a good understanding of trigonometric identities and how they can help simplify complex integrals.

In the exercise example, we integrated the function $ \cos(4-x) $, which is simplified using trigonometric identities and substitutions.

We transformed the integral using the substitution $ u = 4-x $ which altered the limits to coordinates in terms of trigonometric functions.
This, in turn, simplified the integral to manageable terms involving basic trigonometric integrations.

The step-wise use of both substitution and integration by parts with trigonometric functions showcases how various integral calculus techniques can work together to solve complex problems. Mastery of these concepts allows students to approach a wider range of problems in calculus.

Problem 3

Problem 4

Other exercises in this chapter

Problem 3

All the integrals are improper and converge. Explain in each case why the integral is improper, andevaluate each integral. $$ \int_{0}^{\infty} \frac{2}{1+x^{2}

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Problem 3

In Problems , use long division to write $f(x)$ as a sum of a polynomial and a proper rational function.$$ f(x)=\frac{3 x^{3}+5 x-2 x^{2}-2}{x^{2}+1} $$

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Problem 4

Evaluate the indefinite integral by making the given substitution. $$ \int 4 x^{3}\left(4+x^{4}\right)^{1 / 3} d x, \text { with } u=4+x^{4} $$

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Problem 4

Use the midpoint rule to approximate each integral with the specified value of $n$. $\int_{0}^{\pi / 2} \sin x d x, n=4$

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