Question 14 of 33intermediate🔧 ApplyShort Answer2 marks

Find the integrals of the functions in Exercises 1 to 22: sin⁡\sin 3x cos⁡\cos 4x

Correct Answer

To find the integral of sin 3x cos 4x, we use the trigonometric identity for the product of sine and cosine functions.

Using the identity: sin x cos y = ½[sin(x + y) + sin(x - y)]

We get: sin 3x cos 4x = ½[sin(3x + 4x) + sin(3x - 4x)] = ½[sin 7x + sin(-x)] = ½[sin 7x - sin x]

Therefore, ∫ sin 3x cos 4x dx = ½[∫ sin 7x dx - ∫ sin x dx] = ½[-(1/7)cos 7x + cos x] + C = -(1/14)cos 7x + (1/2)cos x + C

Exercise: EXERCISE 7.3 | Q: 2 | (Page 19)
For More Understanding

Explanation

This question requires using the product-to-sum trigonometric identity sin x cos y = ½[sin(x+y) + sin(x-y)], which is demonstrated in the textbook context for similar problems like ∫ sin 2x cos 3x dx. The key steps are: (1) Apply the identity to convert the product into a sum/difference of sines, (2) Integrate each term separately using the standard integral ∫ sin(ax) dx = -(1/a)cos(ax) + C, and (3) Combine results with the constant of integration.

Solution Steps

  1. Step 1: Apply identity sin x cos y = ½[sin(x+y) + sin(x-y)] to get sin 3x cos 4x = ½[sin 7x - sin x]

  2. Step 2: Write the integral as ∫ sin 3x cos 4x dx = ½[∫ sin 7x dx - ∫ sin x dx]

  3. Step 3: Integrate: ∫ sin 7x dx = -(1/7)cos 7x and ∫ sin x dx = -cos x

  4. Step 4: Combine: = ½[-(1/7)cos 7x + cos x] + C = -(1/14)cos 7x + (1/2)cos x + C

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