Question #0c43a

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Answer 1 · 2017-07-16T18:20:37

I got:

#int sin^5 2x dx = -1/10 cos^5 2x + 1/3 cos^3 2x - 1/2 cos 2x + C#

I guess I'd split this using #sin // cos# identities.

#int sin^5 2xdx#

#= int (1 - cos^2 2x)^2 sin 2xdx#

#= int (1 - (-1/2 cdot 2cos 2x)^2)^2 sin 2xdx#

#= int (1 - 4(-1/2 cdot cos 2x)^2)^2 sin 2xdx#

Now, let #u = -1/2cos2x# so that #du = -1/2 cdot -2sin2xdx = sin2xdx#. This gives us:

#= int (1 - 4u^2)^2 du#

#= int 16u^4 - 8u^2 + 1 du#

#= 16/5 u^5 - 8/3 u^3 + u#

Therefore, we get:

#= 16/5 (-1/2 cos 2x)^5 - 8/3 (-1/2 cos 2x)^3 + (-1/2 cos 2x)#

#= color(blue)(-1/10 cos^5 2x + 1/3 cos^3 2x - 1/2 cos 2x + C)#