How do you find the indefinite integral of $int x(cos(8x))^2dx$ ?

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Answer 1 · 2015-10-17T14:25:12

$I = x^2/4+1/32xsin16x+1/512cos16x+C$

$I=int x(cos8x)^2dx$

Integratioin By Parts:

$int udv = uv - int vdu$

$u=x => du=dx$

$(cos8x)^2=cos^2 8x = (1+cos16x)/2$

$dv= (cos8x)^2dx$

$v=int (cos8x)^2dx = 1/2 int (1+cos16x)dx$

$v=1/2(x+1/16sin16x)$

$I = x/2(x+1/16sin16x) - 1/2int (x+1/16sin16x)dx$

$I = x/2(x+1/16sin16x) - 1/2(x^2/2-1/16*1/16cos16x)+C$

$I = 1/2(x^2+1/16xsin16x-x^2/2+1/256cos16x)+C$

$I = x^2/4+1/32xsin16x+1/512cos16x+C$

How do you find the indefinite integral of int x(cos(8x))^2dxint x(cos(8x))^2dx?