can someone show me step by step in solving these problems?
$\displaystyle \int{x^2 sin4x^3} {dx} $
and
$\displaystyle \int{sin^3 2x cos^2 2x} {dx} $
thx in advance of ur help
Neither of these are integration by part, but are u substitions.
For the first one use the u sub $\displaystyle u=4x^3 \implies du=12x^2dx$ to get
$\displaystyle \int{x^2 sin4x^3} {dx} =\frac{1}{12}\int \sin(u)du$
For the 2nd one you need a trig identity...
$\displaystyle \int{\sin^3 2x \cos^2 2x} {dx} =\int \sin(2x)[1-\cos^2(2x)]\cos^2(2x)dx=$
$\displaystyle \int \sin(2x)\cos^2(2x)dx-\int \sin(2x)\cos^4(2x)dx$
Now just let $\displaystyle u=\cos(2x)$ and you are off to the races.