[table][br][tr][br][td][b]Factual Inquiry Questions[/b][/td][br][td][b]Conceptual Inquiry Questions[/b][/td][br][td][b]Debatable Inquiry Questions[/b][/td][br][/tr][br][tr][br][td]1. What is the formula for integration by parts, and how is it derived from the product rule for differentiation?[/td][br][td]1. Why does the integration by parts method work, and what does it tell us about the relationship between differentiation and integration?[/td][br][td]1. Is integration by parts always the most efficient method for integrating products of functions? Why or why not?[/td][br][/tr][br][tr][br][td]2. In the process of using integration by parts, what are the typical choices for the functions [math]u[/math] and [math]dv[/math]?[/td][br][td]2. How does the choice of u and dv affect the complexity and solvability of the integral when using integration by parts?[/td][br][td]2. Could there be alternative methods to integration by parts that have not been discovered yet that could simplify certain integrals?[/td][br][/tr][br][tr][br][td]3. Can you provide a step-by-step example of applying integration by parts to the integral, [math]\int xe^xdx[/math]?[/td][br][td]3. In what scenarios is it necessary to apply integration by parts multiple times, and how can we identify such cases?[/td][br][td]3. How does the application of integration by parts in theoretical mathematics differ from its application in practical problem-solving, such as in engineering or physics?[/td][br][/tr][br][/table][br]

[b][u]Repeated use of integration by parts - Mini-investigation[/u][/b][br][br]Consider the integral, [math]I=\int e^xsin(x)dx[/math] :[br][br]Step 1[br]Implement integration by parts on [math]I[/math] by selecting one function for differentiation and the other for integration. [Note: The expression obtained after integration will include a new integral that cannot be immediately solved.][br] [br]Step 2[br]Apply integration by parts once more to the new integral derived from the previous step. Ensure that you continue differentiating and integrating the same functions as in the initial step [i.e., if the exponential function was initially differentiated, continue to do so; similarly, if the trigonometric function was initially differentiated, continue with that function].[br] [br]Step 3[br]Reflect on the formulae derived for [math]I[/math] after the second application of integration by parts in step 2.[br] [br]Step 4[br]Determine a method to express [math]I[/math] that eliminates the inclusion of any integrals.[br] [br][br][br]Use your findings, to calculate [math]I=\int e^xcos(x)dx[/math].[br] [br]Identify the what characteristics of exponential and trigonometric functions that facilitate the integration of their product.[br][br]

Practice questions 1-6[br][br]Section A - Short response style questions 7-18[br]Challenging questions 11,13,14[br][br]Section B - Long response style questions 19-21

[b]Deduction of the integration by parts formula}[/b][br][br]It is already known how to take the derivative of a multiplication by the product rule: assume[br][br][math]y=uv[/math] [br][br]hence[br][br][math]\frac{dy}{dx}=\frac{d(uv)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}.[/math][br][br]Rearranging this rule:[br][br][math]u\frac{dv}{dx}=\frac{d(uv)}{dx}-v\frac{du}{dx}[/math]. [br][br]Proceed to integrate both sides:[br][br][math]\int u\frac{dv}{dx}dx=\int\frac{d(uv)}{dx}dx-\int v\frac{du}{dx}dx[/math]. [br][br]The initial term on the right-hand side becomes simplified as it involves integrating an already differentiated function.[br][br][math]\int\frac{dv}{dx}dx=uv-\int v\frac{du}{dx}dx[/math]. [br][br]This equation is recognized as the integration by parts formula.