Which method is typically used for integrating products of functions like x sin(x)?

Which method is typically used for integrating products of functions like x sin(x)?

• A. Substitution
• B. Integration by Parts
• C. Partial Fractions
• D. Trigonometric Substitution

Answer: (B)

When integrating products of functions, such as \( x \sin(x) \), the method that is typically most effective is integration by parts. This technique arises from the product rule for differentiation and is useful for integrating functions that are products of algebraic and transcendental functions or two different types of functions. The formula for integration by parts is given by: \[ \int u \, dv = uv - \int v \, du \] In the case of \( x \sin(x) \), you can choose \( u = x \) (which is simple to differentiate) and \( dv = \sin(x) \, dx \) (which integrates neatly). This allows you to find \( du \) and \( v \) and then apply the integration by parts formula. The reason this method is suited for this kind of product is that it breaks the integration process into simpler parts that can be easier to handle. You first compute \( uv \) and then the integral of \( v \, du \), which often leads to a solvable integral. Other methods like substitution, partial fractions, and trigonometric substitution are more suitable for different kinds of integrals. For instance, substitution is great for a single function present

When integrating products of functions, such as ( x \sin(x) ), the method that is typically most effective is integration by parts. This technique arises from the product rule for differentiation and is useful for integrating functions that are products of algebraic and transcendental functions or two different types of functions.

The formula for integration by parts is given by:

[

\int u , dv = uv - \int v , du

]

In the case of ( x \sin(x) ), you can choose ( u = x ) (which is simple to differentiate) and ( dv = \sin(x) , dx ) (which integrates neatly). This allows you to find ( du ) and ( v ) and then apply the integration by parts formula.

The reason this method is suited for this kind of product is that it breaks the integration process into simpler parts that can be easier to handle. You first compute ( uv ) and then the integral of ( v , du ), which often leads to a solvable integral.

Other methods like substitution, partial fractions, and trigonometric substitution are more suitable for different kinds of integrals. For instance, substitution is great for a single function present