what is the anti-derivative of cos^5x

cos^5θ = cos^4θ cosθ dθ

= (1 - sin^2θ)^2 cosθ dθ
= 1 - 2sin^2θ + sin^4θ cosθ dθ

Now, note that if u = sinθ then du = cosθ dθ and we have

1 - 2u^2 + u^4 du

Easy peasie lemon-squeezy:

u - 2/3 u^3 + 1/5 u^5 + C
sinθ - 2/3 sin^3θ + 1/5 sin^5θ + C

To find the antiderivative of cos^5x, you can use a method called integration by substitution. Here's the step-by-step process:

Step 1: Start by rewriting the expression cos^5x as (cos^4x) * cosx.

Step 2: Use the power-reducing identity for cosine, which states that cos^2x = (1 + cos2x)/2. Applying this identity twice, we get cos^4x = ((1 + cos2x)/2)^2.

Step 3: Simplify the expression ((1 + cos2x)/2)^2 by expanding and squaring. This yields ((1 + 2cos2x + cos^22x)/4).

Step 4: Now, rewrite the integral using the above substitutions: ∫((1 + 2cos2x + cos^22x)/4) * cosx dx.

Step 5: Split the integral into three separate integrals: 1/4 * ∫cosx dx + 1/4 * 2∫cos2x * cosx dx + 1/4 * ∫cos^22x * cosx dx.

Step 6: Calculate each integral separately:
- The first integral ∫cosx dx is simply sinx.
- For the second integral ∫cos2x * cosx dx, you can use the substitution u = 2x to simplify it: 1/2 * ∫cosu du. This integral can be solved easily, resulting in (1/2) * sinu + C = (1/2) * sin(2x) + C.
- Finally, for the third integral ∫cos^22x * cosx dx, substitute v = 2x to simplify it: (1/8) * ∫(1 + cos2v) cosv dv. Again, you can solve this integral using the power-reducing identity for cosine: (1/8) * (v + (sin2v)/2) + C = (1/8) * (2x + (sin(4x))/2) + C = (1/4) * x + (sin(4x))/16 + C.

Step 7: Sum up the three integrals: (1/4) * sinx + (1/2) * sin(2x) + (1/4) * x + (sin(4x))/16 + C.

Therefore, the antiderivative of cos^5x is (1/4) * sinx + (1/2) * sin(2x) + (1/4) * x + (sin(4x))/16 + C, where C is the constant of integration.