Question

Use integration by parts to evaluate \(\int x \cos x \, d x\) with \(u=x\) and \(d v=\cos x d x\).

Short answer

Answer: The integral of \(x\cos{x}\) is \(x\sin{x} + \cos{x} + C\), where C is the integration constant.

Step-by-step solution

Identify u and dv

We have been given the two components: \(u = x\) and \(dv = \cos{x} dx\).

Find du and v

We need to find the derivative of \(u\) with respect to \(x\) which gives us \(du\): $$du = \frac{d}{dx}(x) dx = dx$$ We also need to find the integral of \(dv\) which gives us \(v\): $$v = \int \cos{x} dx = \sin{x} + C$$

Apply the integration by parts formula

Now we need to apply the integration by parts formula: $$\int u \, dv = uv - \int v \, du$$ Plugging in our expressions for \(u, v, du,\) and \(dv\), we get: $$\int x \cos{x} dx = x(\sin{x} + C) - \int (\sin{x} + C) dx$$

Evaluate the integral

Now evaluate the integral on the right side of the equation (ignoring the integration constant C): $$\int x\cos{x} dx = x\sin{x} - \int \sin{x} dx = x\sin{x} + \cos{x} + C$$ Therefore, $$\int x\cos{x} dx = x\sin{x} + \cos{x} + C$$

Key concepts

Definite Integrals

Definite integrals are used to compute the area under a curve, between two points on the x-axis. Unlike indefinite integrals, which provide a family of function solutions, definite integrals yield a specific numerical value.

For example, when calculating the area under the curve of \( f(x) \), between \( a \) and \( b \) on the x-axis, the definite integral is written as \( \int_a^b f(x) \, dx \). The calculation requires identifying the antiderivative (or integral) of \( f(x) \) and then evaluating it at the upper and lower bounds of \( a \) and \( b \) respectively. The final result is found by subtracting the value of the antiderivative at \( a \) from the value at \( b \).

In the case of the given exercise, if we were to find the definite integral of \( x \cos x \) from \( a \) to \( b \) after using integration by parts, we would evaluate the boundary conditions after integrating. This would result in \( [x\sin{x} + \cos{x}]_a^b \), which computes the exact area under the curve between \( a \) and \( b \) without any constant of integration.

Integration Techniques

Integration by parts is a powerful technique used in calculus to integrate products of functions. It is based on the product rule for differentiation and provides a way to transform a complex integral into simpler, more manageable parts.

The basic formula for integration by parts is \( \int u \, dv = uv - \int v \, du \), where \( u \) is a function of \( x \) and \( dv \) is another function of \( x \) expressed in differential form. The process involves four key steps:

• Choose \( u \) and \( dv \) from the integral you are dealing with.
• Find \( du \) by differentiating \( u \) and find \( v \) by integrating \( dv \).
• Plug \( u \)\, \( dv \)\, \( du \)\, and \( v \) into the integration by parts formula.
• Simplify and integrate any remaining terms as necessary.

Following these steps systematically can simplify many otherwise difficult integrals, and it is particularly useful when integrating polynomials multiplied by trigonometric, exponential, or logarithmic functions, as shown in the original exercise with \( x \cos x \).

Calculus

Calculus is a branch of mathematics that deals with rates of change (differential calculus) and accumulation of quantities (integral calculus). The two main concepts are the derivative, representing the rate of change, and the integral, which measures the accumulation.

While derivatives are used to determine the slope of a function, integrals are used to find areas under curves, volumes, and other quantities that accumulate over an interval. For students, understanding calculus involves a mastery of limits, functions, derivatives, integrals, and the Fundamental Theorem of Calculus, which connects derivatives and integrals in a profound way.

The exercise provided relates to integral calculus, specifically to integration techniques such as integration by parts, which helps break down complicated products of functions into simpler integrals. Calculus is not only about applying formulas but also about understanding the principles behind them and how they relate to the real world, such as in physics, engineering, economics, and natural sciences.