Question

Find$\int \sin (ax+b)\cos (ax+b)dx$

• A. $\frac{1}{4a}$ $cos(ax+b)+c$
• B. $\frac{-1}{2a}$ $cos2(ax+b)+c$
• C. $\frac{-1}{4a}$ $cos2(ax+b)+c$
• D. $\frac{1}{2a}$ $cos(ax+b)+c$

Answer 1

Answer: (C)

$\frac{-1}{4a}$ $cos2(ax+b)+c$

$\int sin(ax+b)cos(ax+b)dx=1/2\int sin(2ax+2b)dx\left(\frac{1}{2}\right)\left(\frac{-cos(2ax+2b)}{2a}\right)+C=\left(\frac{-1}{4a}\right)cos2(ax+b)+C$