`intsin^(3)(2x+1)dx=?`

Options

A. `(1)/(8)sin^(4)(2x+1)+C`
B. `(1)/(2)cos(2x+1)+(1)/(3)cos^(3)(2x+1)+C`
C. `-(1)/(2)cos(2x+1)+(1)/(6)cos^(3)(2x+1)+C`
D. none of these

Answer

C

Explanation

Put `(2x+1)=t anddx=(1)/(2)dt`.
`:." "I=(1)/(2)intsin^(3)tdt=(1)/(2)int(1-cos^(2)t)sintdt`
`=(1)/(2)int(1-u^(2))(-du)=(1)/(2)int(u^(2)-1)du=(1)/(2)[(u^(3))/(3)-u]+C`, where cos t=u
`=(1)/(6)u^(3)-(1)/(2)u+C=(1)/(6)cos^(3)t-(1)/(2)cost+C=(1)/(6)cos^(3)(2x+1)-(1)/(2)cos(2x+1)+C`.