Question

интеграл cos^2*3x+1/2

Answer 1

$int frac{cos ^{2}3x+1 }{2} , dx =[cos ^{2} alpha = frac{1+cos2 alpha }{2} ]= int frac{ frac{1+cos6x}{2} +1 }{2} } , dx = \ = int{ {( frac{1+cos6x}{4}} +frac{1 }{2}) } , dx = int{ {( frac{cos6x}{4}} +frac{3 }{4})} , dx= frac{1}{4cdot 6} int( cos6xd(6x) +intfrac{3 }{4}dx= \ = frac{sin6x}{24}+ frac{3}{4}x+C$