Question

How do you integrate `int (x+1)sin(x^2+2x+3)` using substitution?

Answer 1

`-1/2cos(x^2+2x+3)+C`.

Explanation:

Le us subst. `x^2+2x+3=t`, so, `(2x+2)dx=2(x+1)dx=dt`.

Therefore, `I=int(x+1)sin(x^2+2x+3)dx`

`=1/2int(2x+2)sin (x^2+2x+3)dx=1/2intsint dt=-1/2cost`.

Hence, `I=-1/2cos(x^2+2x+3)+C`.