How do you find the integral of #x * cos^2 (x)dx#?

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Answer 1 · 2018-04-27T08:20:03

#I=x^2/4+1/8[2xsin2x+cos2x]+c#

We know that,

#color(red)((1)cos^2theta=(1+cos2theta)/2#

#color(blue)((2)intcosAxdx=(sinAx)/A+c#

#color(violet)((3)intsinAxdx=(-cosAx)/A+c#

Here,

#I=intxcos^2xdx#

#=intx((1+cos2x)/2)dx...toApply(1)#

#=1/2intxdx+1/2intxcos2xdx#

#"Using "color(blue)"Integration by Parts"# in second integral

#I=1/2x^2/2+1/2[x int cos2xdx-intd/(dx)(x)intcos2xdx)dx]#

#=x^2/4+1/2[x((sin2x)/2)-int(1)((sin2x)/2)dx]...toApply(2)#

#=x^2/4+1/4[xsin2x-intsin2xdx]#

#=x^2/4+1/4[xsin2x-((-cos2x)/2)]+c...toApply(3)#

#I=x^2/4+1/8[2xsin2x+cos2x]+c#