Question 19 of 33intermediate🔧 ApplyShort Answer2 marks

Find the integrals of the functions in Exercises 1 to 22: sin⁡\sin 4x sin⁡\sin 8x

Correct Answer

∫ sin 4x sin 8x dx = (sin 4x)/8 - (sin 12x)/24 + C

Exercise: EXERCISE 7.3 | Q: 7 | (Page 19)
For More Understanding

Explanation

This question requires using the trigonometric identity for product of sines, similar to the identity shown in the context for sin x cos y. The identity sin A sin B = ½[cos(A-B) - cos(A+B)] transforms the product into a difference of cosines, making integration straightforward.

Solution Steps

  1. Step 1: Use the identity sin A sin B = ½[cos(A-B) - cos(A+B)]

  2. Step 2: Apply to sin 4x sin 8x = ½[cos(4x-8x) - cos(4x+8x)] = ½[cos 4x - cos 12x]

  3. Step 3: Integrate: ∫ sin 4x sin 8x dx = ½∫(cos 4x - cos 12x) dx

  4. Step 4: = ½[(sin 4x)/4 - (sin 12x)/12] + C = (sin 4x)/8 - (sin 12x)/24 + C

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