Integral of cos^5x dx?

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Answer 1 · 2018-06-07T22:07:36

#1/80sin(5x)+5/48sin(3x)+5/8sin(x)+C#

#intcos^5(x)dx#

If you know the complex exponential form of #cos(x)# then you could proceed using the following method:

#cos^5(x)=((e^(ix)+e^(-ix))/(2))^5#

Now, expand this with the binomial theorem to get:

#=1/32(e^(i5x)+5e^(i3x)+10e^(ix)+10e^(-ix)+5e^(-3ix)+e^(-5ix))#

Now collect the terms like so:

#=1/32({e^(i5x)+e^(-i5x)}+5{e^(i3x)+e^(-i3x)}+10{e^(-ix)+e^(ix)})#

#=1/16({e^(i5x)+e^(-i5x)}/2+5{e^(i3x)+e^(-i3x)}/2+10{e^(-ix)+e^(ix)}/2)#

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

So from this it may follow that:

#intcos^5(x)dx=int1/16cos(5x)+5/16cos(3x)+5/8cos(x)dx#

Now integrate to get:

#1/80sin(5x)+5/48sin(3x)+5/8sin(x)+C#