Question 16 of 33intermediate🔧 ApplyShort Answer2 marks

Find the integrals of the functions in Exercises 1 to 22: sin⁡3\sin^3(2x + 1)

Correct Answer

∫ sin³(2x + 1) dx = -½ cos(2x + 1) + ⅙ cos³(2x + 1) + C

Exercise: EXERCISE 7.3 | Q: 4 | (Page 19)
For More Understanding

Explanation

The question asks for the integral of sin³(2x + 1). Using the substitution method shown in the textbook for sin³x, we express sin³(2x+1) as sin²(2x+1)·sin(2x+1) = (1-cos²(2x+1))·sin(2x+1). Then substituting cos(2x+1) = t, we get -sin(2x+1)·2dx = dt, or -sin(2x+1)dx = dt/2. The integral becomes -½∫(1-t²)dt, which integrates to -½t + t³/6 + C. Substituting back gives the final answer.

Solution Steps

  1. Step 1: Express sin³(2x + 1) = sin²(2x + 1)·sin(2x + 1) = (1 - cos²(2x + 1))·sin(2x + 1)

  2. Step 2: Substitute cos(2x + 1) = t, so -sin(2x + 1)·2dx = dt, giving -sin(2x + 1)dx = dt/2

  3. Step 3: ∫ sin³(2x + 1) dx = -½∫(1 - t²) dt = -½[t - t³/3] + C

  4. Step 4: Substitute back: = -½ cos(2x + 1) + ⅙ cos³(2x + 1) + C

PreviousNext