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what is the antiderivative of cos^5x

math 
Steve,
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 lemonsqueezy:
u  2/3 u^3 + 1/5 u^5 + C
sinθ  2/3 sin^3θ + 1/5 sin^5θ + C