Question 15 of 33intermediate🔧 ApplyShort Answer2 marks

Find the integrals of the functions in Exercises 1 to 22: cos⁡\cos 2x cos⁡\cos 4x cos⁡\cos 6x

Correct Answer

116sin⁡\frac{1}{16} \sin 4x + 132sin⁡\frac{1}{32} \sin 8x + 148sin⁡\frac{1}{48} \sin 12x + x4\frac{x}{4} + C

Exercise: EXERCISE 7.3 | Q: 3 | (Page 19)
For More Understanding

Explanation

To integrate the product of three cosine functions, we use the product-to-sum identity 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A \cos B = \cos(A+B) + \cos(A-B), which is the cosine counterpart to the sine identity provided in the context. We simplify the expression by combining terms two at a time. First, we combine cos⁡4x\cos 4x and cos⁡6x\cos 6x, then multiply by cos⁡2x\cos 2x and expand. The resulting terms are either linear cosines (which integrate to sines) or a cos⁡2\cos^2 term, which is integrated using the identity cos⁡2θ=1+cos⁡2θ2\cos^2 \theta = \frac{1+\cos 2\theta}{2} found in the context.

Solution Steps

  1. Step 1: Apply the identity 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A \cos B = \cos(A+B) + \cos(A-B) to the pair cos⁡4xcos⁡6x\cos 4x \cos 6x:

    cos⁡4xcos⁡6x=12[cos⁡(10x)+cos⁡(2x)]\cos 4x \cos 6x = \frac{1}{2}[\cos(10x) + \cos(2x)]
  2. Step 2: Substitute back into the integral:

    I=∫cos⁡2x⋅12[cos⁡10x+cos⁡2x] dx=12∫(cos⁡2xcos⁡10x+cos⁡22x) dxI = \int \cos 2x \cdot \frac{1}{2}[\cos 10x + \cos 2x] \, dx = \frac{1}{2} \int (\cos 2x \cos 10x + \cos^2 2x) \, dx
  3. Step 3: Simplify the first term using the identity again and the second term using cos⁡2x=1+cos⁡2x2\cos^2 x = \frac{1+\cos 2x}{2}:

    cos⁡2xcos⁡10x=12[cos⁡(12x)+cos⁡(8x)]\cos 2x \cos 10x = \frac{1}{2}[\cos(12x) + \cos(8x)] cos⁡22x=1+cos⁡4x2\cos^2 2x = \frac{1 + \cos 4x}{2}
  4. Step 4: Combine all terms inside the integral:

    I=12∫[12(cos⁡12x+cos⁡8x)+12(1+cos⁡4x)] dx=14∫(cos⁡12x+cos⁡8x+1+cos⁡4x) dxI = \frac{1}{2} \int [\frac{1}{2}(\cos 12x + \cos 8x) + \frac{1}{2}(1 + \cos 4x)] \, dx = \frac{1}{4} \int (\cos 12x + \cos 8x + 1 + \cos 4x) \, dx
  5. Step 5: Integrate term by term:

    I=14[sin⁡12x12+sin⁡8x8+x+sin⁡4x4]+CI = \frac{1}{4} [\frac{\sin 12x}{12} + \frac{\sin 8x}{8} + x + \frac{\sin 4x}{4}] + C
  6. Step 6: Final Answer:

    I=sin⁡12x48+sin⁡8x32+x4+sin⁡4x16+CI = \frac{\sin 12x}{48} + \frac{\sin 8x}{32} + \frac{x}{4} + \frac{\sin 4x}{16} + C