Question

How do you find the antiderivative of `x * cos^2 (x)`?

Answer 1

`x^2/4+(xsin2x)/4+(cos2x)/8+C.`

Explanation:

Let `I=intxcos^2xdx.` We use Trgo. Identity `: cos^2x=(1+cos2x)/2`

`:. I=int{x(1+cos2x)}/2dx =1/2intxdx+1/2intxcos2xdx=x^2/4+1/2I_1,`

where, `I_1=intxcos2xdx,` & to evaluate this, we are going to use the Rule of Integration by Parts , given by,

`intuvdx=uintvdx-int[(du)/dx*intvdx]dx.`

We, in this Rule, will take `u=x`, and, `v=cos2x`. Hence,

`I_1=x*intcos2xdx-int[d/dx(x)*intcos2xdx]dx,`

`=x*(sin2x)/2-int1*(sin2x)/2dx,=x/2*sin2x-1/2*(-cos2x)/2.`

`:. I_1=x/2*sin2x+(cos2x)/4.`

Therefore, `I=x^2/4+1/2{(xsin2x)/2+(cos2x)/4}`
`=x^2/4+(xsin2x)/4+(cos2x)/8+C.`

Answer 2

`int x*cos^2 x d x=(2x(sin2x+x)+cos2x)/8+C`

Explanation:

`int x*cos^2 x d x=?`

`cos 2x=cos^2x-sin^2x" ; so "sin^2 x=1-cos^2x`

`cos2x=cos^2 x-(1-cos^2 x)`

`cos2x=cos^2 x-1+cos^2 x=2cos^2 x-1`

`cos2x+1=2cos^2 x`

`cos^2 x=1/2(cos2x+1)`

`int x* cos^2 x d x=int x*1/2(cos2x+1) d x`

`int x*cos^2 x d x=1/2 int x*(cos2x+1) d x`

`int x*cos^2 x d x=1/2[color(red)(int x*cos2x* d x)+color(green)(int x* d x)]`

`color(red)(int x*cos2x*d x)=?" ; "`

`x=u" ; "d x=d u`

`cos 2x d x=d v" ; " v=1/2 sin 2x`

`color(red)(int x*cos2 x d x=u*v-int v*d u)`

`color(red)(int x*cos2 x d x=1/2*x*sin 2x-int 1/2*sin2x d x`

`color(red)(int x*cos2 x d x=1/2x*sin 2x+1/2*1/2*cos2x`
`color(red)(int x*cos2 x d x=1/2 x*sin2x+1/4cos2x`

`color(green)(int x*d x=?)" ; "`

`color(green)(int x*d x=1/2 x^2)`

`int x*cos^2 x d x=1/2[1/2x*sin2x+1/4cos2x+1/2x^2]`

`int x*cos^2 x d x=1/4x*sin2x+1/8*cos2x+1/4*x^2`

`int x*cos^2 x d x=(2xsin2x+2x^2+cos2x)/8+C`

`int x*cos^2 x d x=(2x(sin2x+x)+cos2x)/8+C`