What is the integral of int x * cos^2 (x)dx?

1 Answer
Jun 22, 2016

int x* cos^2 x* d x=color(red)(1/4*x^2)+color(green)(1/4*x*sin 2x+1/8cos 2x+C)

Explanation:

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

cos(x+x)=cos x*cos x-sin x*sin x

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

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

cos 2x=cos^2 x-1+cos^2 x" ; "cos2x=2cos^2 x-1

cos2x+1=2 cos^2 x" ; "1/2 cos 2x+1/2=cos^2 x

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

int x* cos^2 x* d x=int 1/2*x*cos 2x* d x+int1/2*x*d x

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

color(red)(1/2int x*d x=1/2*1/2*x^2=1/4*x^2)

color(green)(1/2* int x*cos 2x*d x)=?

x=u" ; " d x=d u

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

int u*d v=u*v-int v*d u

color(green)(1/2* int x*cos 2x*d x=1/2[1/2*x*sin 2 x-int 1/2*sin2x*d x]

color(green)(1/2* int x*cos 2x*d x=1/4x*sin 2x+1/2*1/2*1/2 cos 2x]

color(green)(1/2* int x*cos 2x*d x=1/4*x*sin 2x+1/8cos 2x

int x* cos^2 x* d x=color(red)(1/4*x^2)+color(green)(1/4*x*sin 2x+1/8cos 2x+C)