Woooowww... that problem is a freaking beast of a problem... that took sooooo many identities to do that. I thought at first glance it was easy, but it isn't a simple u-substitution like I thought at first glance... In a test atmosphere I couldn't pull that problem off.
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[Int] (1-cos2x)/2 (sin2x/2)^2 dx
1/8 [Int] (1-cos2x)sin^2(2x)dx
1/8[Int]sin^2(2x)dx-1/8[Int]sin^2(2x)cos2xdx
= -1/16(1/3sin^3(2x) + 1/16[Int](1-cos4x)dx
= 1/16[x - 1/4sin4x - 1/3sin^3(2x)] + c
Woooowww... that problem is a freaking beast of a problem... that took sooooo many identities to do that. I thought at first glance it was easy, but it isn't a simple u-substitution like I thought at first glance... In a test atmosphere I couldn't pull that problem off.
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