1. ## trig powers integration

$\displaystyle \int (8x+sinx)^2dx$

$\displaystyle \int (sinx^2+16xsinx+64x^2)dx$

$\displaystyle \int (\frac{1-cos2x}{2}+16xsinx+64x^2)dx$

this is where I get stuck

2. Originally Posted by silencecloak
$\displaystyle \int (8x+sinx)^2dx$

$\displaystyle \int (sinx^2+16xsinx+64x^2)dx$

$\displaystyle \int (\frac{1-cos2x}{2}+16xsinx+64x^2)dx$

this is where I get stuck
You can split your last sum to get

$\displaystyle \int \frac{1-cos(2x)}{2}dx + \int 16xsin(x) dx + \int 64x^2 dx$

$\displaystyle \int \frac{1-cos(2x)}{2}dx = \frac{1}{2}\int -\frac{cos(2x)}{2} = -\frac{1}{4}sin(2x)$ (eq1)

the next one needs the substitution rule:

$\displaystyle u = 16x ; u' = 16$
$\displaystyle v' = sin(x) ; v = cos(x)$

$\displaystyle \int uv^{\prime} = uv - \int vu^{\prime}$

$\displaystyle 16xcos(x) - \int (16cos(x))dx = 16(x - sin(x))$ (eq2)

the last one is simple

$\displaystyle \int 64x^2 = \frac{64}{3}x^3$ (eq3)

Add our equations together (not forgetting the constant of integration):

$\displaystyle [-\frac{1}{4}sin(2x)] + [16(x - sin(x))] + \frac{64}{3}x^3 + C$

3. Ive never seen some of those techniques...

4. I'm not sure that is correct.

Webassign (online math) won't take that, but I could be wrong

Either way