In number 4,since it's integral vdu isn't it supposed to be 1/2 sin 2x*2x since du is 2x

The integration by parts formula is $\int u \, dv = uv - \int v \, du$.

For , we set:
$u = x^2$
$du = 2x \, dx$
$dv = \cos(2x) \, dx$
$v = \int \cos(2x) \, dx = \frac{1}{2}\sin(2x)$

Now, let's substitute these into the $\int v \, du$ part of the formula:

$$\int v \, du = \int \left(\frac{1}{2}\sin(2x)\right) (2x \, dx)$$

Here, $v = \frac{1}{2}\sin(2x)$ and $du = 2x \, dx$. We are integrating their product.

To simplify this integral:

$$\int \frac{1}{2}\sin(2x) (2x \, dx) = \int \left(\frac{1}{2} \cdot 2\right) x\sin(2x) \, dx$$ $$= \int 1 \cdot x\sin(2x) \, dx$$ $$= \int x\sin(2x) \, dx$$

So, the original step was correct:

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

The $2x$ from $du$ is multiplied with $v$ inside the integral, and then the entire expression is integrated.