Question 25 of 33intermediate🔧 ApplyShort Answer2 marks

Integrate the functions in Exercises 1 to 22: x sin⁡\sin 3x

Correct Answer

-x3cos⁡\frac{x}{3} \cos 3x + 19sin⁡\frac{1}{9} \sin 3x + C

Exercise: EXERCISE 7.6 | Q: 2 | (Page 39)
For More Understanding

Explanation

The question requires the integration of a product of two functions, which is solved using the Integration by Parts method. Although the specific formula for Integration by Parts is not explicitly stated in the provided context chunks, the integration of trigonometric functions like sin⁡3x\sin 3x (yielding −13cos⁡3x-\frac{1}{3}\cos 3x) is supported by the patterns seen in the context (e.g., ∫sin⁡3x dx\int \sin 3x \, dx in Chunk 2). The solution applies the standard ILATE rule to choose xx as the first function and sin⁡3x\sin 3x as the second function.

Solution Steps

  1. Step 1: Identify the integral I=∫xsin⁡3x dxI = \int x \sin 3x \, dx. Using Integration by Parts (∫uv dx=u∫v dx−∫(dudx∫v dx)dx)\left( \int u v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx \right), choose the first function u=xu = x (algebraic) and second function v=sin⁡3xv = \sin 3x (trigonometric).

  2. Step 2: Integrate the second function: ∫sin⁡3x dx=−13cos⁡3x\int \sin 3x \, dx = -\frac{1}{3} \cos 3x.

  3. Step 3: Apply the formula: I=x(−13cos⁡3x)−∫(1)(−13cos⁡3x)dxI = x \left( -\frac{1}{3} \cos 3x \right) - \int (1) \left( -\frac{1}{3} \cos 3x \right) dx.

  4. Step 4: Simplify and integrate the remaining term: I=−x3cos⁡3x+13∫cos⁡3x dx=−x3cos⁡3x+13(13sin⁡3x)+CI = -\frac{x}{3} \cos 3x + \frac{1}{3} \int \cos 3x \, dx = -\frac{x}{3} \cos 3x + \frac{1}{3} \left( \frac{1}{3} \sin 3x \right) + C.

  5. Step 5: Write the final answer: I=−x3cos⁡3x+19sin⁡3x+CI = -\frac{x}{3} \cos 3x + \frac{1}{9} \sin 3x + C.

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