Integral cos(2 x) cos(4 x) dx Use the trigonometric identity cos(alpha) cos(beta) = 1/2 (cos(alpha-beta)+cos(alpha+beta)), where alpha = 2 x and beta = 4 x: = 1/2 integral (cos(2 x)+cos(6 x)) dx Integrate the sum term by term: = 1/2 integral cos(2 x) dx+1/2 integral cos(6 x) dx For the integrand cos(2 x), substitute u = 2 x and du = 2 dx: = 1/4 integral cos(u) du+1/2 integral cos(6 x) dx For the integrand cos(6 x), substitute s = 6 x and ds = 6 dx: = 1/12 integral cos(s) ds+1/4 integral cos(u) du The integral of cos(s) is sin(s): = (sin(s))/12+1/4 integral cos(u) du The integral of cos(u) is sin(u): = (sin(s))/12+(sin(u))/4+constant Substitute back for s = 6 x: = (sin(u))/4+1/12 sin(6 x)+constant Substitute back for u = 2 x: = 1/4 sin(2 x)+1/12 sin(6 x)+constant Which is equal to: = 1/12 (3 sin(2 x)+sin(6 x))+constant

∫(2x - 1)^2 dx

Let u = 2x - 1

du/2 = dx

(1/2)∫(u)^2 du=

u^3/6 + C=

(1/6)(2x-1)^3+C

Answer b)

b) 1/6(2x-1)^3+C

b

∫(2x - 1)^2 dx

u = 2x - 1

du/2 = dx

1/2*∫u^2 du

1/8*(2x - 1)^(3) + C