integral (cos^3(6x))

cos^3(6x) = cos(6x) * cos^2(6x); cos^2(6x) = 1-sin^2(6x); therefore cos^3(6x) =cos(6x)(1-sin^2(6x)) = cos(6x) - cos(6x)*sin^2(6x) Now you use the linearity of the integral operator: \[intcos(6x)dx;let u =6x --> du/dx = 6, so dx = du/6, thus we have (1/6)\intcos(u) du = (1/6)sin(u) +c1...but remember that u = 6x, so our actual solution is (1/6)sin(6x) +c1]

now for the other part: int of cos(6x)*sin^2(6x)...another substitution-this time let u = sin(6x) that implies that du/dx = (1/6)cos(6x) and that dx = 6du/cos(6x)... so we now have: int of cos(6x)*u^2du/cos(6x)...(cos(6x) cancels out and we're left with:) int of u^2 du = (1/3)u^3 +c2 (remember u = sin(6x) so: (1/3)sin^3(6x) +c2... thus our total answer is this: (1/6)sin(6x) +c1 + (1/3)sin^3(6x) +c2 (let c1 +c2 = k since constant +constant =constant) then final solution: (1/6)sin(6x) + (1/3)sin^3(6x) + k

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