Question

Integration of sin 3 (2x + 1) dx

Answer 1

first let u = 2x + 1 --> du/dx = 2 --> dx = du/2 

Thus, 
int (sin^3(2x +1) dx ) = int( sin^3(u) du/2) = (1/2) * int ( sin^3(u) du) 
= (1/2)*int( sin(u)sin^2(u) du) 
= (1/2)*int(sin(u)(1 - cos^2(u)) du) 
= (1/2) [ int (sin(u) du) - int(sin(u)cos^2(u)du) ] 
first integral simple = -cos(u) 
second integral, let w = cos(u) --> dw/du = -sin(u) --> du = -dw/sin(u) 
Thus, 
=(1/2)[ -cos(u) - int(sin(u)w^2 -dw/sin(u) ) ] 
=(1/2)[ -cos(u) + int( w^2 dw) ] 
=(1/2)[ -cos(u) + w^3/3 ] + C; C= integration constant 
now resub back in 
=(1/2)[-cos(u) + (1/3)cos^3(u)] + C 
=(1/2)[-cos(2x + 1) + (1/3)cos^3(2x + 1)] + C 

Hope this helps,