How do you find the integral of x * cos^2 (x)dxxcos2(x)dx?

1 Answer
Apr 27, 2018

I=x^2/4+1/8[2xsin2x+cos2x]+cI=x24+18[2xsin2x+cos2x]+c

Explanation:

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