need help with a integral (sin^2(t)cos^4(t)), I can split it up into: ((1-cos(2t))/2)((1+cos(2t))/2)^2, but the solution then takes out the constants and make it so its 1/8 infront of the integral, how do you get 1/8? I can get 1/4 but not 1/8

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need help with a integral (sin^2(t)cos^4(t)), I can split it up into: ((1-cos(2t))/2)((1+cos(2t))/2)^2, but the solution then takes out the constants and make it so its 1/8 infront of the integral, how do you get 1/8? I can get 1/4 but not 1/8

Mathematics
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The way I would do it is to apply\[\sin^2t=1-\cos^2t\]them use the double angle formula\[\cos^2t=\frac12(1+\cos(2t))\]
you wanted to know how we get 1/8 ? [(1-...)/2 ] [(1+...)/2]^2 = [1/2-.....] [1/4+ ...] = [1/8 +....]
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ah, i see what your saying now
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it was taking the second part and squaring it and expanding it that was confusing but when I expanded it i forgot about what happened to my 2 on the bottom yes, thank you
welcome ^_^

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