Question 4 of 33intermediate🔧 ApplyShort Answer2 marks

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

Correct Answer

∫ sin⁴x dx = 3x/8 - sin(2x)/4 + sin(4x)/32 + C

Exercise: EXERCISE 7.3 | Q: 10 | (Chapter: 19)
For More Understanding

Explanation

The question requires evaluating the integral of sin⁴x. Following the approach shown in Example 29 for sin²x, we use the identity sin²x = (1 - cos 2x)/2. Squaring this gives sin⁴x = [(1 - cos 2x)/2]². Expanding and applying the identity cos²2x = (1 + cos 4x)/2, we express sin⁴x as a sum of integrable terms. The final integration is done term by term.

Solution Steps

  1. Step 1: Using sin²x = (1 - cos 2x)/2, write sin⁴x = [(1 - cos 2x)/2]² = (1 - 2cos 2x + cos²2x)/4

  2. Step 2: Apply cos²2x = (1 + cos 4x)/2 to get sin⁴x = (3 - 4cos 2x + cos 4x)/8

  3. Step 3: Integrate term by term: ∫sin⁴x dx = ∫(3/8)dx - ∫(cos 2x)/2 dx + ∫(cos 4x)/8 dx

  4. Step 4: Final answer: 3x/8 - sin(2x)/4 + sin(4x)/32 + C

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