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July 1, 2015

July 1, 2015

Posted by **.....................................** on Tuesday, December 20, 2011 at 2:52pm.

- math -
**Steve**, Tuesday, December 20, 2011 at 3:37pmcos^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