We know that,
color(red)((1)cos^2theta=(1+cos2theta)/2(1)cos2θ=1+cos2θ2
color(blue)((2)intcosAxdx=(sinAx)/A+c(2)∫cosAxdx=sinAxA+c
color(violet)((3)intsinAxdx=(-cosAx)/A+c(3)∫sinAxdx=−cosAxA+c
Here,
I=intxcos^2xdxI=∫xcos2xdx
=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