First, I would simplify the squared term:
INT x^2 + 2xsin3x + sin^2(3x) dx
Now this integration has three parts: x^2 (which is easy), 2xsin3x (which needs by-parts), sin^2(3x) (which needs trig identity).
INT x^2 dx + 2*INT xsin3x dx + INT sin^2(3x) dx
For INT x^2 dx:
INT x^2 dx
= 1/3*x^3
For INT 2xsin3x dx:
Let u = 2x <--> du = 2dx
Let dv = sin3x dx <--> v = -1/3*cos3x
=> -(2/3)xcos3x - 2/3*INT cos3x dx
= -(2/3)xcos3x - (2/9)sin3x
For INT sin^2(3x) dx:
Let sin^2(3x) = 1/2(1 - cos6x) <-- this is the half-angle formula for sin(3x) after squaring both sides.
1/2*INT 1 - cos6x dx
= 1/2*(x - 1/6*sin6x)
= (1/2)x - (1/12)sin6x
Recombining these integrations, we get:
(1/3)x^3 - (2/3)xcos3x - (2/9)sin3x + (1/2)x - (1/12)sin6x + C
Good question. I skipped that step, but I'll explain it.
We have:
dv = 1/sqrt(2 + 3x) dx
To get v, we need to integrate both sides:
INT dv = INT 1/sqrt(2 + 3x) dx
The integration of dv is easy. The integration of 1/sqrt(2 + 3x) requires an additional substitution.
Let n = 2 + 3x <--> dn = 3 dx --> dx = 1/3 dn
1/3*INT 1/sqrt(n) dn
= 1/3*INT (n)^(-1/2) dn
= 1/3*(2/1)*n^(1/2)
= (2/3)sqrt(2 + 3x)
Therefore:
v = (2/3)sqrt(2 + 3x)
Anytime I show something like:
Let dv = f(x) dx <--> v = F(x)
Where F(x) is the antiderivative of f(x), I recommend you try to figure out how I went from f(x) dx to F(x). Try doing the integration yourself. It should improve your ability to do integrations.
Let u = sinx^(1/3) <--> du = 1/3*x^(-2/3)*sinx^(1/3) dx
Let dv = dx <--> v = x
=> xsinx^(1/3) - (1/3)*INT x^(-2/3)sinx^(1/3) dx
Let w = x^(1/3) <--> dw = (1/3)x^(-2/3) dx --> dx = 3*x^(2/3) dw
=> xsinx^(1/3) - INT sinw dw
= xsinx^(1/3) + cosw + C
= xsinx^(1/3) + cosx^(1/3) + C