How do you evaluate the integral of #int (sin^4x)dx#?

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Answer 1 · 2016-01-17T20:03:45

Definitively the fastest way to do that, is to use euler formula

#sin^4(x) = (1/(2i)(e^(ix)-e^(-ix)))^4#

#(1/(2i))^4 = (1/2^4)# because #i^4 = 1#

expand with binomial theorem

#e^(4ix)-4e^(3ix)e^(-ix)+6e^(2ix)*e^(-2ix)-4e^(ix)*e^(-3ix)+e^(-4ix)#

#=1/2^4((e^(4ix)+e^(-4ix))-4(e^(2ix)+e^(-2ix))+6)#

#=1/2^4(2cos(4x) - 8cos(2x) + 6)#

so

#int sin^4(x) dx = 1/2^4int2cos(4x)-8cos(2x)+6dx#

which is

#1/2^4[1/2sin(4x)-4sin(2x)+6x]#