integral and differential problem from my lecture 1

i got this problem from my lecture,
1. integral equation, i need to know step by step to got the answer
$$\int y^3 \sin(2x) dx$$
my friends said the answer is $-y^3 \cos^2(x)$, but i don't know how to get this.

1. differential equation,
a. $\frac{Dm}{dy} (3y^2 + y \sin (2xy))$
b. $\frac{Dn}{dx}( 6xy + x \sin (2xy))$
my friend said again two of them have same answer, it said 6y + sin 2xy + 2xy cos 2xy
but once again i don't know how to get thia,

someone can tell me step by step please, i really appreciate it, thanks a lot.

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Those aren't differential equations. A differential equation needs a derivative of something present, like $y^\prime=y$. – icurays1 Dec 12 '12 at 16:23
What is $m$ and $n$? – AD. Dec 12 '12 at 16:34

If I'm interpreting your question correctly, you want to find:

$$\int y^3\sin(2x)dx,$$

$$\frac{\partial }{\partial y}(3y^2+y\sin(2xy))$$and

$$\frac{\partial }{\partial x}(6xy+x\sin(2xy))$$

The first is an integral with respect to $x$, so we can pull the $y^3$ out: $$\int y^3\sin(2x)dx=y^3\int\sin(2x)dx=y^3\left(-\frac{1}{2}\cos(2x)+C\right)$$ so your friend is incorrect.

For the second, we differentiate with respect to $y$ so $x$ is treated as a constant:

$$\frac{\partial }{\partial y}(3y^2+y\sin(2xy))=6y+\frac{\partial }{\partial y}(y\sin(2xy))=6y+\sin(2xy)+2xy\cos(2xy)$$ so your friend is correct. For the third, we differentiate with respect to $x$ so $y$ is treated as a constant:

$$\frac{\partial}{\partial x}(6xy+x\sin(2xy))=6y+\sin(2xy)+2yx\cos(2xy)$$so that one is correct as well.

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