Question

Integration of sin (3x+2) dx​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\int\sin(3x+2)\,dx$

$\mapsto\,\tt{Put\,\,\,3x+2=t}$

$\mapsto\,\tt{3\,dx=dt}$

So,

$\displaystyle\int\sin(t)\,\dfrac{dt}{3}$

$\displaystyle=\dfrac{1}{3}\int\sin(t)\,dt$

$\displaystyle=\dfrac{1}{3}\cdot\big\{-\cos(t)\big\}+C$

$\displaystyle=-\dfrac{1}{3}\cos(t)+C$

$\displaystyle=-\dfrac{1}{3}\cos(3x+2)+C$