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What is the integral of #cos^5(x)#?

1 Answer
Dec 16, 2014

#=sinx+(sin^5x)/5-2/3*sin^3x+c#, where #c# is a constant

Explanation :

#=int(cos^5x) dx#

From trigonometric identity, which is

#cos^2x+sin^2x=1#, #=>cos^2x=1-sin2x#

#=int(cos^4x)*cos(x) dx#

#=int(cos^2x)^2*cos(x) dx#

#=int(1-sin^2x)^2*cos(x) dx# .. #(i)#

let's assume #sinx = t#, #=> (cosx) dx= dt#

substituting this in the #(i)#, we get

#=int(1-t^2)^2dt#

Now using expansion of #(1-y)^2=1+y^2-2y#, yields,

#=int(1+t^4-2t^2)dt#

#=intdt+intt^4dt-2intt^2dt#

#=t+t^5/5-2*t^3/3+c#, where #c# is a constant

#=t+t^5/5-2/3*t^3+c#, where #c# is a constant

now substituting #t# back gives,

#=sinx+(sinx)^5/5-2/3*(sinx)^3+c#, where #c# is a constant

#=sinx+(sin^5x)/5-2/3*sin^3x+c#, where #c# is a constant