What is #intcos^3xdx#?

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Answer 1 · 2017-12-10T04:03:02

#intcos^3xdx=sinx-1/3sin^3x+C#

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#intcos^3xdx=intcos^2xcosxdx=int(1-sin^2x)cosxdx=#

#int(cosx-six^2xcosx)dx=#

#intcosxdx-intsin^2xcosxdx=sinx-intsin^2xcosxdx=sinx-I#

To solve the last integral, we can use #u#-substitution:

Let #u=sinx#, then #du=cosxdx#

#I=intu^2du=1/3u^3+C#

Now, we substitute back:

#I=1/3sin^3x+C#

Therefore,

#intcos^3xdx=sinx-1/3sin^3x+C#