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)#