Question 33 of 33intermediate🔧 ApplyShort Answer2 marks

Integrate the functions in Exercises 1 to 22: exe^{x} (sin⁡\sin x + cos⁡\cos x)

Correct Answer

Let I = ∫ exe^{x} (sin x + cos x) dx

This integral is of the type ∫ exe^{x} [f(x) + f'(x)] dx, where f(x) = sin x and f'(x) = cos x.

Using the formula: ∫ exe^{x} [f(x) + f'(x)] dx = exe^{x} · f(x) + C

Therefore, I = exe^{x} sin x + C

Exercise: EXERCISE 7.6 | Q: 16 | (Page 39)
For More Understanding

Explanation

This question relates to section 7.6.1 of the textbook which covers integrals of the type ∫ exe^{x} [f(x) + f'(x)] dx. The textbook shows that when integrating by parts, this type of integral simplifies to exe^{x} · f(x) + C. Here, sin x + cos x can be recognized as f(x) + f'(x) where f(x) = sin x, since the derivative of sin x is cos x.

Solution Steps

  1. Step 1: Identify f(x) = sin x, so f'(x) = cos x

  2. Step 2: Recognize the integral is of the form ∫ exe^{x} [f(x) + f'(x)] dx

  3. Step 3: Apply the formula: ∫ exe^{x} [f(x) + f'(x)] dx = exe^{x} · f(x) + C

  4. Step 4: Substitute f(x) = sin x to get the answer: exe^{x} sin x + C

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