Find the integrals of the functions in Exercises 1 to 22: 2x
We evaluate ∫ cos⁴ 2x dx. Using cos² θ = (1+cos2θ)/2, we have cos⁴ 2x = (cos² 2x)² = [(1+cos4x)/2]² = (1/4)(1 + 2cos4x + cos² 4x). Integrating: ∫1 dx = x, ∫ cos4x dx = (1/4) sin4x, and ∫ cos² 4x dx = (1/2)(x + (1/8) sin8x) = x/2 + (1/16) sin8x. Thus, ∫ cos⁴ 2x dx = (1/4)[x + (1/2) sin4x + x/2 + (1/16) sin8x] + C = (3x/8) + (1/8) sin4x + (1/64) sin8x + C.
Explanation
The question asks to integrate cos⁴ 2x. The textbook context provides the identity cos² x = (1+cos2x)/2 and shows integration of cos² x. By applying this identity twice, we reduce the power and integrate termwise. This is a standard method for integrating even powers of cosine.
Solution Steps
Step 1: Use identity cos² θ = (1+cos2θ)/2 to write cos² 2x = (1+cos4x)/2.
Step 2: Square to get cos⁴ 2x = (1/4)(1 + 2cos4x + cos² 4x).
Step 3: Integrate each term: ∫1 dx = x, ∫ cos4x dx = (1/4) sin4x, and ∫ cos² 4x dx = (1/2)(x + (1/8) sin8x) = x/2 + (1/16) sin8x.
Step 4: Combine: I = (1/4)[x + (1/2) sin4x + x/2 + (1/16) sin8x] + C = (3x/8) + (1/8) sin4x + (1/64) sin8x + C.